Limits & Continuity

Problem 601

Find limx+2x+5x27x+3\lim _{x \rightarrow+\infty} \frac{2 x+5}{x^{2}-7 x+3}
Add your answer Integer, decimal, or E notation allowed

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Problem 602

Find limx4x2x12x4\lim _{x \rightarrow 4} \frac{x^{2}-x-12}{x-4}
Add your answer Integer, decimal, or E notation allowed

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Problem 603

Determine where f(x)\mathrm{f}(\mathrm{x}) is continuous. f(x)=sin12xf(x)=\sin ^{-1} 2 x (A) 1/2-1 / 2 and 1/21 / 2 (B) 1 and - 1/21 / 2 (C) -1 and 0 (D) -1 and -3

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Problem 604

Is the function f such that f(x)=1xf(x)=\frac{1}{x} for x>0\mathrm{x}>0 and f(0)=0\mathrm{f}(0)=0 continuous over [0,1][0,1] ? (A) no (B) yes (C) maybe (D) none of the answer

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Problem 605

Evaluate the limit: limx04x+497x=\lim _{x \rightarrow 0} \frac{\sqrt{4 x+49}-7}{x}= \square

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Problem 606

Exercice 5 Considérons la fonction ff définie par: f(x)=x22x+3x+3f(x)=\frac{x^{2}-2 x+3}{x+3}
1. Déterminer le domaine de définition DfD_{f} de ff.
2. Calculer les limites aux bornes de ff.
3. Interpréter les limites aux bornes réelles.
4. Déterminer les a,ba, b et cc tels que f(x)=ax+b+cx+3f(x)=a x+b+\frac{c}{x+3}.
5. En déduire que la droite (D)(D) d'équation (D):y=x5(D): y=x-5 est une asymptote oblique pour (Cf)\left(C_{f}\right).

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Problem 607

For the following limits, (a) Determine the Indeterminate form. (b) Compute the limit using L'Hopital's Rule (i) limx0cos(mx)cos(nx)x2\lim _{x \rightarrow 0} \frac{\cos (m x)-\cos (n x)}{x^{2}} where n& m\mathrm{n} \& \mathrm{~m} are even interger constant.
Hint: Utilitze the Unit Circle for when n or m=2,4,6,\mathrm{m}=2,4,6, \ldots (ii) limx0+sin(x)ln(x)\lim _{x \rightarrow 0^{+}} \sin (x) \ln (x) (iii) limx0+xex1\lim _{x \rightarrow 0^{+}} \frac{x}{e^{x}-1}

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Problem 608

ابحث في اتصال ق ( س) عند س = 1 ، وغند س = ب من جهة اليسار وايضا ابحث الاتصال على الفترة ( . ، )

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Problem 609

Теорема 3.2.12. Любой отрезок [a,b]R[a, b] \subsetneq \mathbb{R} конечной длины a<ba<b, является компактным множеством.
Доказательство. Доказывать будем от противного. Долустим, что существует такое покрытие {ϑα}αAγ\left\{\vartheta_{\alpha}\right\}_{\alpha \in A}^{\gamma} открытыми множествами из R\mathbb{R} для отрезка I=[a,b]I=[a, b], что из него нельзя выбрать конечное подпокрытие.
Итак, пусть IαAUαI \subseteq \bigcup_{\alpha \in A} \mathcal{U}_{\alpha} и из этого покрытия нельзя выбрать конечное подпокрытие, которое бы покрыло II. Разобьём II пополам т.е представим его так: [a,b]=[a,a+b2][a+b2,b].[a, b]=\left[a, \frac{a+b}{2}\right] \cup\left[\frac{a+b}{2}, b\right] .
По условию II нельзя покрыть конечным числом множеств из {Uα}αA\left\{U_{\alpha}\right\}_{\alpha \in A}, тогда хотя бы один из полученных отрезков, обозначим его через I1I_{1} тоже нельзя покрыть конечным числом множеств из покрытия {α}αA\left\{\bigcup_{\alpha}\right\}_{\alpha \in A}. Иначе бы оба полученных отрезка покрывались бы конечным числом множества из покрытия, а тогда и I покрывался бы конечным числом множеств из этого покрытия, что и показывала бы компактность II.
Разобьём теперь отрезок I1I_{1} аналогичным образом на два равных отрезка. Так как I1I_{1} нельзя покрыть конечным числом множеств из покрытия {α}αA\left\{थ_{\alpha}\right\}_{\alpha \in A}, то найдётся хотя бы один, скажем, I2I_{2}, из только что полученных, который тоже нельзя покрыть конечным числом множеств. Будем повторять эту процедуру каждый раз, пусть Ik=[ak,bk],k1I_{k}=\left[a_{k}, b_{k}\right], k \geq 1. В результате мы получаем бесконечную цепь вложенных друг в друга отрезков II1I2,I \supsetneq I_{1} \supsetneq I_{2} \supsetneq \ldots,
каждый из которых нельзя покрыть конечным числом элементов множества {Uα}αA\left\{U_{\alpha}\right\}_{\alpha \in A}. Более того, их длины строго уменьшаются (каждый из отрезков по длине в два раза меньше, чем предыдущий). Тогда по Лемме о вложенных отрезках (Лемма 1.5.6), существует такая точка cIc \in I, что ck1Ikc \in \bigcap_{k \geq 1} I_{k}, которая есть предел для последовательности их концов; limkak=c=limnbn.\lim _{k \rightarrow \infty} a_{k}=c=\lim _{n \rightarrow \infty} b_{n} .
Тогда для любого ε>0\varepsilon>0 найдётся такой номер NN, что при kNk \geq N все ak,bk(cε,c+ε)a_{k}, b_{k} \in(c-\varepsilon, c+\varepsilon), а по построению это значит, что и все Ik(cε,c+ε),kNI_{k} \subseteq(c-\varepsilon, c+\varepsilon), k \geq N.
С другой стороны, так как имеется покрытие {ϑα}αA\left\{\vartheta_{\alpha}\right\}_{\alpha \in A} этого отрезка II, то найдётся хотя бы одно открытое множество UαU_{\alpha} такое, что сUαс \in U_{\alpha}, а так как оно открытое, то для точки сс можно найти ε\varepsilon-окрестность (cε,c+ε)Uα(c-\varepsilon, c+\varepsilon) \subseteq U_{\alpha}.
Таким образом, мы получаем, что для всех kNk \geq N есть включения Ik(cε,c+ε)Uα,I_{k} \subseteq(c-\varepsilon, c+\varepsilon) \subseteq U_{\alpha},
но это значит, что все отрезки IkI_{k} покрываются всего одним открытым множеством UαU_{\alpha}, что противоречит построению отрезков IkI_{k}, т.е. такое построение невозможно, что и означает компактность отрезка II. 58
Теперь у нас всё готово, чтобы описать компактные множества в R\mathbb{R}.

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Problem 610

lim[x](x2+3x+2+x+1)\lim [x \rightarrow \infty]\left(\sqrt{x^{2}+3 x+2}+x+1\right)

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Problem 611

limx0[x7(sinxx5120+x36x)]\lim _{x \rightarrow 0}\left[x^{-7}\left(\sin x-\frac{x^{5}}{120}+\frac{x^{3}}{6}-x\right)\right]

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Problem 613

5) limx16xx2+16\lim _{x \rightarrow-\infty}-\frac{16 x}{x^{2}+16}

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Problem 614

limx+1+xxlnx\lim _{x \rightarrow+\infty} 1+x-x \ln x

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Problem 615

Find the secant line slope for f(x)=x311f(x)=\sqrt[3]{x}-11 at x1=750x_1=750 and x2=6x_2=6.

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Problem 616

Find the limit: limx0(cot(6x)sin(2x))\lim _{x \rightarrow 0}(\cot (6 x) \cdot \sin (2 x)).

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Problem 617

Find the left-hand and right-hand limits of f(x)=sin(x)f(x) = \sin(x) as xx approaches 1, and the limit as xx approaches -1.

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Problem 618

Find the limit: limx7x3343x7\lim _{x \rightarrow 7} \frac{x^{3}-343}{x-7}.

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Problem 619

Find the limit of f(x+h)f(x)h\frac{f(x+h)-f(x)}{h} as h0h \to 0 for f(x)=2x+3f(x) = 2x + 3.

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Problem 620

Find the limit: limx(5x2)(7x22)8x3(x210)\lim _{x \rightarrow \infty} \frac{(5-x^{2})(7 x^{2}-2)}{-8 x^{3}(x^{2}-10)}. State DNE if infinite.

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Problem 621

Find the limit: limx24x39x22x3x22x+1\lim _{x \rightarrow \infty} \frac{-24 x^{3}-9 x^{2}}{2 x^{3}-x^{2}-2 x+1}. State if it DNE.

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Problem 622

Find the limit as xx approaches infinity: limx(17x3)(x8)(3x21)(2x2+3)\lim _{x \rightarrow \infty} \frac{(1-7 x^{3})(x-8)}{(3 x^{2}-1)(2 x^{2}+3)}. State if it does not exist (DNE).

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Problem 623

Find the limit: limx(17x3)(x8)(3x21)(2x2+3)\lim _{x \rightarrow \infty} \frac{(1-7 x^{3})(x-8)}{(3 x^{2}-1)(2 x^{2}+3)}. State DNE if infinite.

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Problem 624

Find the limit as xx approaches infinity: limx(1+10x3)(4+x2)(10x3)(x23)\lim _{x \rightarrow \infty} \frac{(1+10 x^{3})(4+x^{2})}{(10-x^{3})(x^{2}-3)}. State DNE if infinite.

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Problem 625

Find the limit: limx40x5+15x218x327x5+46x2\lim _{x \rightarrow \infty} \frac{40 x^{5}+15 x^{2}}{18 x^{3}-27 x^{5}+4-6 x^{2}}. State if it is infinite (DNE).

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Problem 626

Find the limit: limx(1+10x3)(4+x2)(10x3)(x23)\lim _{x \rightarrow \infty} \frac{(1+10 x^{3})(4+x^{2})}{(10-x^{3})(x^{2}-3)}. State if it DNE.

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Problem 627

Find the limit: limx(7+5x2)(4x37)(2x3+9)(1+9x2)\lim _{x \rightarrow \infty} \frac{(7+5 x^{2})(4 x^{3}-7)}{(2 x^{3}+9)(1+9 x^{2})}. State if it DNE.

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Problem 628

Find the limit: limx2x2+x333x10x2+6\lim _{x \rightarrow \infty} \frac{\sqrt{-2 x^{2}+x^{3}-33 x}}{10 x^{2}+6}. State DNE if infinite.

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Problem 629

Find the limit: limx36+36x+5x290x629x3+2\lim _{x \rightarrow \infty} \frac{36+36 x+5 x^{2}}{90 x^{6}-29 x^{3}+2}. State if it DNE.

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Problem 630

Find the limit: limx42x2+16x105x+4x4+4x2\lim _{x \rightarrow \infty} \frac{\sqrt{42 x^{2}+16 x^{10}}}{5 x+4 x^{4}+4 x^{2}}. State DNE if infinite.

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Problem 631

Find the limit: limx2x2+x333x10x2+6\lim _{x \rightarrow \infty} \frac{\sqrt{-2 x^{2}+x^{3}-33 x}}{10 x^{2}+6}. State if it DNE.

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Problem 632

Find the limit as xx approaches infinity: limx42x2+16x105x+4x4+4x2\lim _{x \rightarrow \infty} \frac{\sqrt{42 x^{2}+16 x^{10}}}{5 x+4 x^{4}+4 x^{2}}. State if it DNE.

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Problem 633

Find the limit as xx approaches infinity: limx5x4+33x564x7310x+8+9x3\lim _{x \rightarrow \infty} \frac{\sqrt[3]{-5 x^{4}+33 x^{5}-64 x^{7}}}{10 x+8+9 x^{3}}. If infinite, state DNE.

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Problem 634

Find the limit: limx5x4+33x564x7310x+8+9x3\lim _{x \rightarrow \infty} \frac{\sqrt[3]{-5 x^{4}+33 x^{5}-64 x^{7}}}{10 x+8+9 x^{3}}. State if it DNE.

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Problem 635

Find the limit: limx5x4+33x564x7310x+8+9x3\lim _{x \rightarrow \infty} \frac{\sqrt[3]{-5 x^{4}+33 x^{5}-64 x^{7}}}{10 x+8+9 x^{3}}. State DNE if infinite.

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Problem 636

Determine the limit: limxx339x20x335log5x\lim _{x \rightarrow \infty} \frac{-x^{33}-9 x^{20}}{x^{33}-5 \log _{5} x}. What does it equal?

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Problem 637

Determine the limit: limx6x91x19+x91\lim _{x \rightarrow \infty} \frac{-6 x^{91}}{x^{19}+x^{91}}. What does it equal?

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Problem 638

Determine the limit: limx7x46lnx6ex\lim _{x \rightarrow \infty} \frac{-7 x^{46}}{\ln x 6 e^{x}}. What does it equal?

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Problem 639

Find the horizontal asymptotes of the function f(x)=1+2e1xf(x)=1+2 e^{\frac{1}{x}}.

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Problem 640

Evaluate the limit: limx07sin(3x)3x\lim _{x \rightarrow 0} \frac{7 \sin (3 x)}{3 x}.

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Problem 641

Find the limit as xx approaches 0: limx07sin(3x)3x\lim _{x \rightarrow 0} \frac{7 \sin (3 x)}{3 x}.

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Problem 642

Evaluate the limit: limx03x+1cos(x)2x\lim_{x \to 0} \frac{-3x + 1 - \cos(x)}{2x}

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Problem 643

(b) limx+,x>01x=0\lim _{x \rightarrow+\infty, x>0} \frac{1}{x}=0.

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Problem 644

(1) (1,0 ponto) Use a Regra de L'Hôspital para calcular os seguintes limites, se for necessário: (a) limx+lnxx3\lim _{x \rightarrow+\infty} \frac{\ln x}{\sqrt[3]{x}} (b) limx0xex+cosx\lim _{x \rightarrow 0} \frac{x}{e^{x}+\cos x} (c) limx0+senxlnx\lim _{x \rightarrow 0^{+}} \operatorname{sen} x \ln x

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Problem 645

6\sqrt{\circ} 6 gokazamb limx(1+1x)x=e\lim _{x \rightarrow \infty}\left(1+\frac{1}{x}\right)^{x}=e

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Problem 646

Use a Regra de L'Hôspital para calcular os seguintes limites, se for necessário: (c) limx0+senxlnx\lim _{x \rightarrow 0^{+}} \operatorname{sen} x \ln x

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Problem 647

Evaluate the limit: limx03sin2(x)4x\lim _{x \rightarrow 0} \frac{3 \sin ^{2}(x)}{4 x}.

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Problem 648

Evaluate the limit: limx04sin(x)7x\lim _{x \rightarrow 0} \frac{-4 \sin (x)}{7 x}.

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Problem 649

Find the horizontal asymptotes of the function f(x)=1+2e1xf(x)=1+2 e^{\frac{1}{x}}.

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Problem 650

Find the horizontal asymptotes of the function f(x)=3x44x55x3x4f(x)=\frac{3 x^{4}-4 x^{5}}{5 x^{3}-x^{4}}.

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Problem 651

Find the horizontal asymptotes for the function f(x)=2e1xf(x)=2 e^{\frac{1}{x}}.

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Problem 652

limx(2x23x25x+3)\lim _{x \rightarrow \infty}\left(\frac{2 x^{2}-3}{x^{2}-5 x+3}\right)

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Problem 653

limx021+cosxsin2x\lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sin ^{2} x}

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Problem 654

\#2. (18 points) Evaluate the following limits: (2a) (6 points) limxxsin(1x)\lim _{x \rightarrow \infty} x \sin \left(\frac{1}{x}\right).

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Problem 655

1. limx3+2x45x\lim _{x \rightarrow-\infty} \frac{3+2^{x}-\infty}{4-5^{x}} is (A) 25-\frac{2}{5} (B) 0 (C) 34\frac{3}{4}

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Problem 656

fonksiyon için δ\delta, sadece ϵ\epsilon sayısına bağlıdır yani x=tx=t den bağımsız olmaktadır. 3.11 Örnek f:(0,1]R,f(x)=1xf:(0,1] \longrightarrow \mathbb{R}, f(x)=\frac{1}{x} fonksiyonunun sürekli ancak düzgün sürekli olmadığım gösteriniz.

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Problem 657

Evaluate the limit: limx1sin(x2+1)x(5x)\lim _{x \rightarrow 1} \frac{\sin \left(x^{2}+1\right)}{\sqrt{x}(5-x)}

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Problem 658

Find the value(s) of xx where the function f(x)f(x) is discontinuous, defined as: f(x)={x+1if x<02cos(πx)if 0x26x2if x>2 f(x) = \begin{cases} x+1 & \text{if } x<0 \\ 2 \cos (\pi x) & \text{if } 0 \leq x \leq 2 \\ 6-x^{2} & \text{if } x>2 \end{cases}

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Problem 659

Evaluate limx0f(x)\lim _{x \rightarrow 0} f(x) given 2cosxf(x)x2+12-\cos x \leq f(x) \leq x^{2}+1.

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Problem 660

The function ff is defined for all xx in the interval 4<x<64<x<6. Which of the following statements, if true, implies that limx5f(x)=17\lim _{x \rightarrow 5} f(x)=17 ? A) There exists a function gg with f(x)g(x)f(x) \leq g(x) for 4<x<64<x<6, and limx5g(x)=17\lim _{x \rightarrow 5} g(x)=17.
B There exists a function gg with g(x)f(x)g(x) \leq f(x) for 4<x<64<x<6, and limx5g(x)=17\lim _{x \rightarrow 5} g(x)=17. (C) There exist functions gg and hh with f(x)g(x)h(x)f(x) \leq g(x) \leq h(x) for 4<x<64<x<6, and limx5g(x)=limx5h(x)=17\lim _{x \rightarrow 5} g(x)=\lim _{x \rightarrow 5} h(x)=17. (D) There exist functions gg and hh with g(x)f(x)h(x)g(x) \leq f(x) \leq h(x) for 4<x<64<x<6, and limx5g(x)=limx5h(x)=17\lim _{x \rightarrow 5} g(x)=\lim _{x \rightarrow 5} h(x)=17.

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Problem 661

Find the limit LL. L=limx52x2+14x+20x+5L=\lim _{x \rightarrow-5} \frac{2 x^{2}+14 x+20}{x+5}

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Problem 662

Mail - JENIFFER MARIE - [-/2 Points] DETAILS MY NOTES LARCALCET8 2.R.043.
Use a graphing utility to graph the function and estimate the limit LL. Use a table to reinforce your conclusion. Then find the limit LL by analytic methods. limx02x+42xL=\begin{array}{l} \lim _{x \rightarrow 0} \frac{\sqrt{2 x+4}-2}{x} \\ L=\square \end{array}
Show My Work (Required)

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Problem 663

Find the limit LL (if it exists). (If an answer does not exist, enter DNE.) limx9x9x281L=\begin{array}{l} \lim _{x \rightarrow 9^{-}} \frac{x-9}{x^{2}-81} \\ L=\square \end{array}
If it does not exist, explain why. The limit does not exist at x=9x=9 because the function approaches different values from the left and right side of 9. The limit does not exist at x=9x=9 because the function value is undefined at x=9x=9. The limit does not exist at x=9x=9 because the function does not approach f(9)f(9) as xx approaches 9 . The limit does not exist at x=9x=9 because the function is not continuous at any xx value. The limit exists at x=9x=9.

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Problem 664

Determine whether f(x)f(x) approaches \infty or -\infty as xx approaches 4 from the left and from the right. f(x)=1(x4)2limx41(x4)2=limx4+1(x4)2=\begin{array}{r} f(x)=\frac{-1}{(x-4)^{2}} \\ \lim _{x \rightarrow 4^{-}} \frac{-1}{(x-4)^{2}}=\square \\ \lim _{x \rightarrow 4^{+}} \frac{-1}{(x-4)^{2}}=\square \end{array} \square \square

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Problem 665

Question 9 Not yet answered fligg quartion
If f(x)={x2xx21 if x>12k3 if x1f(x)=\left\{\begin{array}{ll}\frac{x^{2}-x}{x^{2}-1} & \text { if } x>1 \\ 2 k-3 & \text { if } x \leq 1\end{array}\right. is continuous at x=1x=1, then k=k= 12\frac{1}{2} 34\frac{3}{4} 1 74\frac{7}{4} 74\frac{-7}{4}

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Problem 666

Find the limit LL (if it exists). (If an answer does not exist, enter DNE.) limx5x5x5L=\begin{array}{l} \lim _{x \rightarrow 5^{-}} \frac{|x-5|}{x-5} \\ L=\square \end{array}
If it does not exist, explain why. The limit does not exist at x=5x=5 because the function is not continuous at any xx value. The limit does not exist at x=5x=5 because the function approaches different values from the left and right side of 5 . The limit does not exist at x=5x=5 because the function value is undefined at x=5x=5. The limit does not exist at x=5x=5 because the function does not approach f(5)f(5) as xx approaches 5 . The limit exists at x=5x=5.

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Problem 667

Find limx4x2x12x4\lim _{x \rightarrow 4} \frac{x^{2}-x-12}{x-4}

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Problem 668

(B) > (D) 1>2
9. Find lim 1-e* x+0 In (2-e*) (A) 210 (B) 1 (C) 0 (D) The limit does not exist. function given by f(x)=2 graph of fat

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Problem 669

d) limx3x32x2+1x2+1\lim _{x \rightarrow-\infty} \frac{3 x^{3}-2 x^{2}+1}{x^{2}+1}

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Problem 670

Find the limit: limx0sin(6x)sin(2x)\lim _{x \rightarrow 0} \frac{\sin (6 x)}{\sin (2 x)}.

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Problem 671

Find δ\delta such that if x1<δ|x-1|<\delta, then f(x)1<0.2|f(x)-1|<0.2, given f(1)=1f(1)=1, f(1.1)=0.8f(1.1)=0.8, f(1.2)=1.2f(1.2)=1.2.

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Problem 672

Find δ\delta so that if 0<x3<δ0<|x-3|<\delta, then f(x)2<0.5|f(x)-2|<0.5 for the function ff at points (2.6,1.5)(2.6,1.5) and (3.8,2.5)(3.8,2.5).

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Problem 673

Find δ\delta such that if x4<δ|x-4|<\delta, then x2<0.4|\sqrt{x}-2|<0.4 for f(x)=xf(x)=\sqrt{x}.

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Problem 674

Find δ\delta so that if x1<δ|x-1|<\delta, then f(x)1<0.2|f(x)-1|<0.2 using points (.7,1.2) and (1.1,.8).

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Problem 675

Find the limit: limx0sin(4x)2x=\lim _{x \rightarrow 0} \frac{\sin (4 x)}{2 x}=

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Problem 676

Find δ\delta so that if 0<x3<δ0<|x-3|<\delta, then f(x)2<0.5|f(x)-2|<0.5 using the graph of ff.

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Problem 677

Identify which functions are continuous for all real numbers: I. f(x)=x1/3f(x)=x^{1 / 3} II. g(x)=secxg(x)=\sec x III. h(x)=exh(x)=e^{-x} (A) I only (B) I and II only (C) I and III only (D) I, II, and III

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Problem 678

Use the graph of f(x)f(x) to answer the questions.
State the xx-values at which f(x)5f(x)^{5} is discontinuous. Note : List all xx-values and separate multiple \square

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Problem 679

5. limx2x2x\lim _{x \rightarrow 2^{-}} \frac{x}{2-x}

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Problem 680

f(x)={sin(3x)sin(5x) if x0ab if x=0f(x)=\left\{\begin{array}{l} \frac{\sin (3 x)}{\sin (5 x)} \text{ if } x \neq 0 \\ \frac{a}{b} \text{ if } x=0 \end{array}\right.
g(x)=(x3+2x+1)ln(x)g(x)=\left(x^{3}+2 x+1\right) \ln (x)
المطلوب: إيجاد مجموعة تعريف الدوال f(x) f(x) و g(x) g(x) .

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Problem 681

Suppose that lim(x,y)(5,2)f(x,y)=7\lim _{(x, y) \rightarrow(5,2)} f(x, y)=7. What can you say about the value of f(5,2)f(5,2) ? We can say f(5,2)=f(5,2)=\infty. We can say f(5,2)f(5,2) is an open point. We can say lim(x,y)(5,2)f(x,y)=f(5,2)=7\lim _{(x, y) \rightarrow(5,2)} f(x, y)=f(5,2)=7. We can say (x,y)(5,2)f(x,y)=f(5,2)=\underset{(x, y) \rightarrow(5,2)}{ } f(x, y)=f(5,2)=\infty. In general, nothing can be said about the value of f(5,2)f(5,2).
What if ff is continuous? We can say f(5,2)=f(5,2)=\infty. We can say f(5,2)f(5,2) is an open point. We can say (x,y)(5,2)(x, y) \rightarrow(5,2). We can say lim(x,y)(5,2)f(x,y)=f(5,2)=\lim _{(x, y) \rightarrow(5,2)} f(x, y)=f(5,2)=\infty. In general, nothing can be said about the value of f(5,2)f(5,2).

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Problem 682

Determine the set of points at which the function is continuous. f(x,y)={x2y32x2+y2 if (x,y)(0,0)1 if (x,y)=(0,0)f(x, y)=\left\{\begin{array}{ll} \frac{x^{2} y^{3}}{2 x^{2}+y^{2}} & \text { if }(x, y) \neq(0,0) \\ 1 & \text { if }(x, y)=(0,0) \end{array}\right. {(x,y)x>0\{(x, y) \mid x>0 and y>0}y>0\} {(x,y)(x,y)(0,0)}\{(x, y) \mid(x, y) \neq(0,0)\} {(x,y)xR\{(x, y) \mid x \in \mathbb{R} and y0}y \neq 0\} {(x,y)xR\{(x, y) \mid x \in \mathbb{R} and yR}y \in \mathbb{R}\} {(x,y)xy0}\{(x, y) \mid x \cdot y \neq 0\}

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Problem 683

7. Evaluate limx0(2x)tan(πx2)\lim _{x \rightarrow 0}(2-x) \tan \left(\frac{\pi x}{2}\right) limx0(2x)tan(πx2)=(20)tanπ(0)2=(20)tan02=(20)=1\begin{array}{l} \lim _{x \rightarrow 0}(-2-x)^{\tan \left(\frac{\pi x}{2}\right)} \\ =(2-0)^{\tan \frac{\pi(0)}{2}} \\ =(2-0)^{\tan \frac{0}{2}} \\ =(2-0)^{\circ} \\ =1 \end{array}

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Problem 684

Exercice 1. 1) Montrer en utilisant la définition que limx14x+2=6,limx0x3+x+2=2,limx+1x3+1=0\lim _{x \rightarrow 1} 4 x+2=6, \lim _{x \rightarrow 0} x^{3}+x+2=2, \lim _{x \rightarrow+\infty} \frac{1}{x^{3}+1}=0 2) Montrer que les limites limx1E(x),limx0cos1x\lim _{x \rightarrow 1} E(x), \lim _{x \rightarrow 0} \cos \frac{1}{x} n'existent pas.
Exercice 2. Calculer les limites suivantes:
1. limx0tanxsinxx3\lim _{x \rightarrow 0} \frac{\tan x-\sin x}{x^{3}},
2. limxπ2ecosx1xπ2\lim _{x \rightarrow \frac{\pi}{2}} \frac{e^{\cos x}-1}{x-\frac{\pi}{2}},
3. limx0xaE\lim _{x \rightarrow 0} \frac{x}{a} E 12x)-1-2 x),
5. limx+(x2+2x12x)\lim _{x \rightarrow+\infty}\left(\sqrt{x^{2}+2 x-1}-2 x\right), 6. limx+(1+1x)x\lim _{x \rightarrow+\infty}\left(1+\frac{1}{x}\right)^{x},
7. limx(x1x+1)x\lim _{x \rightarrow-\infty}\left(\frac{x-1}{x+1}\right)^{x}.

Exercice 3. Soit f:RRf: \mathbb{R} \longrightarrow \mathbb{R}, une fonction définie par f(x)={x si x1x2+ax+b si x>1f(x)=\left\{\begin{array}{ll} x & \text { si }|x| \leqslant 1 \\ x^{2}+a x+b & \text { si }|x|>1 \end{array}\right.
a,bRa, b \in \mathbb{R}. Déterminer les valeurs de aa et bb pour que ff soit continue sur R\mathbb{R}. Exercice 4. Soit f:RRf: \mathbb{R} \longrightarrow \mathbb{R} la fonction définie par f(x)={1 si xQ0 si Q.f(x)=\left\{\begin{array}{l} 1 \text { si } x \in \mathbb{Q} \\ 0 \text { si } \notin \mathbb{Q} . \end{array}\right.
Montrer que ff est discontinue en tout point de R\mathbb{R}. Exercice 5. Soit ff une fonction définie par f:[a,b][a,b]f:[a, b] \rightarrow[a, b], telle que f(x1)f(x2)<x1x2,x1,x2[a,b],(x1x2).\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right|<\left|x_{1}-x_{2}\right|, \quad \forall x_{1}, x_{2} \in[a, b], \quad\left(x_{1} \neq x_{2}\right) . 1) Montrer que ff est continue sur [a,b][a, b]. 2) Montrer que l'équation f(x)=xf(x)=x admet une solution unique.
Exercice 6. Étudier dans chacun des cas suivants si la fonction ff est prolongeable par continuité sur R\mathbb{R}. 1) f:x1cosxxf: x \longrightarrow \frac{1-\cos \sqrt{|x|}}{|x|}, 2) f:xsin(x)sin(1x)f: x \longrightarrow \sin (x) \cdot \sin \left(\frac{1}{x}\right), 3) f:xx3+1x2+3x+2f: x \longrightarrow \frac{x^{3}+1}{x^{2}+3 x+2}, 4) f:x1cos(1cos1x2)f: x \longrightarrow 1-\cos \left(1-\cos \frac{1}{x-2}\right), 5) f:xsinx4x4f: x \longrightarrow \frac{\sin |x-4|}{x-4}, 6) f:x11x21x2f: x \longrightarrow \frac{1}{1-x}-\frac{2}{1-x^{2}}.

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Problem 685

8. Sketch a graph where f(4)f(4) does not exist but limx4f(x)\lim _{x \rightarrow 4} \quad f(x) does exist.

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Problem 686

For each of the following forms determine whether the following limit type is indeterminate, always has a fixed finite value, or never has a fixed finite value. In the first case answer IND, in the second case enter the numerical value, and in the third case answer DNE. To discourage blind guessing, this problem is graded on the following scale 090-9 correct =0=0 101310-13 correct =.3=.3 14-16 correct =.5=.5 17-19 correct =.7=.7 Note that l'Hospital's rule (in some form) may ONLY be applied to indeterminate forms. \square 1. 11 \cdot \infty
2. π\pi^{-\infty}
3. \infty \cdot \infty
4. 1\infty^{1}
5. \infty^{-\infty}
6. π\pi^{\infty}
7. 00^{-\infty}
8. e\infty^{-e}
9. \infty^{\infty}
10. 101^{0}
11. 1\frac{1}{-\infty} 12.112.1^{\infty}
13. 0\frac{0}{\infty}
14. 00 \cdot \infty
15. 00^{\infty}
16. 0\frac{\infty}{0} 17.0017.0^{0} 18.118.1^{-\infty}

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Problem 687

(13) logba×logcb×logac=\log _{\mathrm{b}} \mathrm{a} \times \log _{\mathrm{c}} \mathrm{b} \times \log _{\mathrm{a}} \mathrm{c}= (a) zero (b) 1 (c) a b c (d) ac (14) Limx2(3a2)=\operatorname{Lim}_{x \rightarrow 2}\left(3 a^{2}\right)= \qquad (a) 3 (b) 12 (c) 3a23 a^{2} (d) 6 (15) Limx0x2xx=\operatorname{Lim}_{x \rightarrow 0} \frac{x^{2}-x}{x}= \qquad (a) zero (b) -1 (c) 11 \cdot (d) Doesn't exist. (16) Limx0x+11x=\operatorname{Lim}_{x \rightarrow 0} \frac{\sqrt{x+1}-1}{x}= \qquad (a) zero (b) 2\sqrt{2} (c) 12\frac{1}{2} (d) as no existence (17) If Limx2x24ax2\operatorname{Lim}_{x \rightarrow 2} \frac{x^{2}-4 a}{x-2} exists, then a=\mathrm{a}= \qquad (a) -1 (b) 1 (c) 2 (d) 4 (18) Limx2x532x38=\operatorname{Lim}_{x \rightarrow 2} \frac{x^{5}-32}{x^{3}-8}= \qquad (a) 4 (b) 53\frac{5}{3} (c) zero (d) 6236 \frac{2}{3}

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Problem 688

(10) limsin(x)\lim \sin (x)

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Problem 689

Find the limit: limx01cosx3sin2x\lim _{x \rightarrow 0} \frac{1-\cos x}{3 \sin ^{2} x}.

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Problem 690

Find the limit: limx0tan(2x)3x=\lim _{x \rightarrow 0} \frac{\tan (2 x)}{3 x}= (A) 13\frac{1}{3} (B) 12\frac{1}{2} (C) 23\frac{2}{3} (D) 2

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Problem 691

Find the limit: limx8x255x2+x3\lim _{x \rightarrow \infty} \frac{8 x^{2}-5}{5 x^{2}+x-3}.

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Problem 692

Find the limit: limx6x5xx4+3\lim _{x \rightarrow-\infty} \frac{6 x^{5}-x}{x^{4}+3}.

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Problem 693

78. If f(x)=ln(x)f(x)=\ln (x) and gg is a differentiable function with domain x>0x>0 such that limxg(x)=\lim _{x \rightarrow \infty} g(x)=\infty and gg^{\prime} has a horizontal asymptote at y=4y=4 then limxf(x)g(x)\lim _{x \rightarrow \infty} \frac{f(x)}{g(x)} is A. 0 B. -4 C. 4 D. nonexistent

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Problem 694

23-28 True-False Determine whether tr - statement is true or false. Explain your answer.
23. If f(x)f(x) is continuous at x=cx=c, then so is f(x)|f(x)|.
24. If f(x)|f(x)| is continuous at x=cx=c, then so is f(x)f(x).
25. If ff and gg are discontinuous at x=cx=c, then so is f+gf+g.
26. If ff and gg are discontinuous at x=cx=c, then so is fgf g.

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Problem 695

limx83p(x)2x153x=?\lim_{x \rightarrow 8^{-}} \frac{3p(x) - 2x}{15 - 3x} = ?
Given that as xx approaches 8 from the left, the value of p(x)p(x) approaches 2, and as xx approaches 8 from the right, the value of p(x)p(x) approaches 3.

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Problem 696

(1 point)
Evaluate the limit limxx2+5x+6x\lim _{x \rightarrow \infty} \sqrt{x^{2}+5 x+6}-x

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Problem 697

Find the limit: limx0(4x4sin(x))=\lim _{x \rightarrow 0}\left(\frac{4}{x}-\frac{4}{\sin (x)}\right)= \square (Enter undefined if the limit does not exist.)
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Problem 698

Evaluate the limit using L'Hospital's rule if necessary. limx0+x9sin(x)\lim _{x \rightarrow 0^{+}} x^{9 \sin (x)}
Answer: \square

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Problem 699

(1 point)
Evaluate the following limits, using L'Hopital's rule if appropriate. limx0+x4ln(x)=\lim _{x \rightarrow 0^{+}} \sqrt[4]{x} \ln (x)= \square Preview My Answers Submit Answers

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Problem 700

Find the limit. Use l'Hospital's Rule where appropriate. limxx2ex\lim _{x \rightarrow-\infty} x^{2} e^{x}
Limit: \square
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