Math / Calculus

QuestionEvaluate the following I. (a) i) $\lim _{x \rightarrow 2}\left\{\frac{x^{2}-7 x+10}{x^{2}-4}\right\}$ ii) $\lim _{x \rightarrow 0}\left\{\frac{\sin 7 x}{x}\right\}$ iii) $\lim _{x \rightarrow \infty}\left\{\frac{2 x^{3}-7 x+3}{x^{2}+2}\right\}$ (b) Use the chain rule to differentiate $y=\left(3 t^{2}+2 t-9\right)^{10}$ .

Studdy Solution

STEP 1

Assumptions
1. We need to evaluate the following limits: i) $\lim _{x \rightarrow 2}\left\{\frac{x^{2}-7 x+10}{x^{2}-4}\right\}$ ii) $\lim _{x \rightarrow 0}\left\{\frac{\sin 7 x}{x}\right\}$ iii) $\lim _{x \rightarrow \infty}\left\{\frac{2 x^{3}-7 x+3}{x^{2}+2}\right\}$
2. We need to differentiate $y=\left(3 t^{2}+2 t-9\right)^{10}$ using the chain rule.

STEP 2

Evaluate the first limit: i) $\lim _{x \rightarrow 2}\left\{\frac{x^{2}-7 x+10}{x^{2}-4}\right\}$
First, factorize the numerator and the denominator: $x^{2} - 7x + 10 = (x - 2)(x - 5)$ $x^{2} - 4 = (x - 2)(x + 2)$

STEP 3

Rewrite the limit using the factorized form: $\lim _{x \rightarrow 2}\left\{\frac{(x - 2)(x - 5)}{(x - 2)(x + 2)}\right\}$

STEP 4

Cancel out the common factor $(x - 2)$ : $\lim _{x \rightarrow 2}\left\{\frac{x - 5}{x + 2}\right\}$

STEP 5

Substitute $x = 2$ into the simplified expression: $\frac{2 - 5}{2 + 2} = \frac{-3}{4}$
So, the first limit is: $\lim _{x \rightarrow 2}\left\{\frac{x^{2}-7 x+10}{x^{2}-4}\right\} = \frac{-3}{4}$

STEP 6

Evaluate the second limit: ii) $\lim _{x \rightarrow 0}\left\{\frac{\sin 7 x}{x}\right\}$
We use the standard limit result $\lim _{u \rightarrow 0}\left\{\frac{\sin u}{u}\right\} = 1$ .

STEP 7

Rewrite the limit using the substitution $u = 7x$ : $\lim _{x \rightarrow 0}\left\{\frac{\sin 7 x}{x}\right\} = \lim _{x \rightarrow 0}\left\{\frac{\sin 7 x}{7x} \cdot 7\right\}$

STEP 8

Apply the standard limit result: $\lim _{x \rightarrow 0}\left\{\frac{\sin 7 x}{7x}\right\} = 1$
Thus, the second limit is: $\lim _{x \rightarrow 0}\left\{\frac{\sin 7 x}{x}\right\} = 7$

STEP 9

Evaluate the third limit: iii) $\lim _{x \rightarrow \infty}\left\{\frac{2 x^{3}-7 x+3}{x^{2}+2}\right\}$
Divide the numerator and the denominator by $x^2$ : $\lim _{x \rightarrow \infty}\left\{\frac{2 x^{3}-7 x+3}{x^{2}+2}\right\} = \lim _{x \rightarrow \infty}\left\{\frac{2 x - \frac{7}{x} + \frac{3}{x^2}}{1 + \frac{2}{x^2}}\right\}$

STEP 10

As $x \rightarrow \infty$ , the terms $\frac{7}{x}$ , $\frac{3}{x^2}$ , and $\frac{2}{x^2}$ approach 0: $\lim _{x \rightarrow \infty}\left\{\frac{2 x - 0 + 0}{1 + 0}\right\} = \lim _{x \rightarrow \infty}\left\{2 x\right\}$
Thus, the third limit is: $\lim _{x \rightarrow \infty}\left\{\frac{2 x^{3}-7 x+3}{x^{2}+2}\right\} = \infty$

STEP 11

Differentiate $y=\left(3 t^{2}+2 t-9\right)^{10}$ using the chain rule.
Let $u = 3 t^{2} + 2 t - 9$ , then $y = u^{10}$ .

STEP 12

Differentiate $y$ with respect to $u$ : $\frac{dy}{du} = 10u^9$

STEP 13

Differentiate $u$ with respect to $t$ : $\frac{du}{dt} = 6t + 2$

STEP 14

Apply the chain rule: $\frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt} = 10u^9 \cdot (6t + 2)$

STEP 15

Substitute $u = 3 t^{2} + 2 t - 9$ back into the expression: $\frac{dy}{dt} = 10(3 t^{2} + 2 t - 9)^9 \cdot (6t + 2)$
Thus, the derivative is: $\frac{dy}{dt} = 10(3 t^{2} + 2 t - 9)^9 \cdot (6t + 2)$

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Studdy Solution

STEP 1

STEP 2

Evaluate the first limit: i) $\lim _{x \rightarrow 2}\left\{\frac{x^{2}-7 x+10}{x^{2}-4}\right\}$
First, factorize the numerator and the denominator: $x^{2} - 7x + 10 = (x - 2)(x - 5)$ $x^{2} - 4 = (x - 2)(x + 2)$

STEP 3

Rewrite the limit using the factorized form: $\lim _{x \rightarrow 2}\left\{\frac{(x - 2)(x - 5)}{(x - 2)(x + 2)}\right\}$

STEP 4

Cancel out the common factor $(x - 2)$ : $\lim _{x \rightarrow 2}\left\{\frac{x - 5}{x + 2}\right\}$

STEP 5

Substitute $x = 2$ into the simplified expression: $\frac{2 - 5}{2 + 2} = \frac{-3}{4}$
So, the first limit is: $\lim _{x \rightarrow 2}\left\{\frac{x^{2}-7 x+10}{x^{2}-4}\right\} = \frac{-3}{4}$

STEP 6

Evaluate the second limit: ii) $\lim _{x \rightarrow 0}\left\{\frac{\sin 7 x}{x}\right\}$
We use the standard limit result $\lim _{u \rightarrow 0}\left\{\frac{\sin u}{u}\right\} = 1$ .

STEP 7

Rewrite the limit using the substitution $u = 7x$ : $\lim _{x \rightarrow 0}\left\{\frac{\sin 7 x}{x}\right\} = \lim _{x \rightarrow 0}\left\{\frac{\sin 7 x}{7x} \cdot 7\right\}$

STEP 8

Apply the standard limit result: $\lim _{x \rightarrow 0}\left\{\frac{\sin 7 x}{7x}\right\} = 1$
Thus, the second limit is: $\lim _{x \rightarrow 0}\left\{\frac{\sin 7 x}{x}\right\} = 7$

STEP 9

STEP 10

STEP 11

Differentiate $y=\left(3 t^{2}+2 t-9\right)^{10}$ using the chain rule.
Let $u = 3 t^{2} + 2 t - 9$ , then $y = u^{10}$ .

STEP 12

Differentiate $y$ with respect to $u$ : $\frac{dy}{du} = 10u^9$

STEP 13

Differentiate $u$ with respect to $t$ : $\frac{du}{dt} = 6t + 2$

STEP 14

Apply the chain rule: $\frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt} = 10u^9 \cdot (6t + 2)$

STEP 15

Substitute $u = 3 t^{2} + 2 t - 9$ back into the expression: $\frac{dy}{dt} = 10(3 t^{2} + 2 t - 9)^9 \cdot (6t + 2)$
Thus, the derivative is: $\frac{dy}{dt} = 10(3 t^{2} + 2 t - 9)^9 \cdot (6t + 2)$

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Studdy Solution

STEP 1

STEP 2

STEP 3

Rewrite the limit using the factorized form: lim⁡x→2{(x−2)(x−5)(x−2)(x+2)} \lim _{x \rightarrow 2}\left\{\frac{(x - 2)(x - 5)}{(x - 2)(x + 2)}\right\} x→2lim​{(x−2)(x+2)(x−2)(x−5)​}

STEP 4

Cancel out the common factor (x−2)(x - 2)(x−2): lim⁡x→2{x−5x+2} \lim _{x \rightarrow 2}\left\{\frac{x - 5}{x + 2}\right\} x→2lim​{x+2x−5​}

STEP 5

STEP 6

Evaluate the second limit: ii) lim⁡x→0{sin⁡7xx}\lim _{x \rightarrow 0}\left\{\frac{\sin 7 x}{x}\right\}limx→0​{xsin7x​}We use the standard limit result lim⁡u→0{sin⁡uu}=1\lim _{u \rightarrow 0}\left\{\frac{\sin u}{u}\right\} = 1limu→0​{usinu​}=1.

STEP 7

Rewrite the limit using the substitution u=7xu = 7xu=7x: lim⁡x→0{sin⁡7xx}=lim⁡x→0{sin⁡7x7x⋅7} \lim _{x \rightarrow 0}\left\{\frac{\sin 7 x}{x}\right\} = \lim _{x \rightarrow 0}\left\{\frac{\sin 7 x}{7x} \cdot 7\right\} x→0lim​{xsin7x​}=x→0lim​{7xsin7x​⋅7}

STEP 8

Apply the standard limit result: lim⁡x→0{sin⁡7x7x}=1 \lim _{x \rightarrow 0}\left\{\frac{\sin 7 x}{7x}\right\} = 1 x→0lim​{7xsin7x​}=1Thus, the second limit is: lim⁡x→0{sin⁡7xx}=7 \lim _{x \rightarrow 0}\left\{\frac{\sin 7 x}{x}\right\} = 7 x→0lim​{xsin7x​}=7

STEP 9

STEP 10

STEP 11

Differentiate y=(3t2+2t−9)10y=\left(3 t^{2}+2 t-9\right)^{10}y=(3t2+2t−9)10 using the chain rule.Let u=3t2+2t−9u = 3 t^{2} + 2 t - 9u=3t2+2t−9, then y=u10y = u^{10}y=u10.

STEP 12

Differentiate yyy with respect to uuu: dydu=10u9 \frac{dy}{du} = 10u^9 dudy​=10u9

STEP 13

Differentiate uuu with respect to ttt: dudt=6t+2 \frac{du}{dt} = 6t + 2 dtdu​=6t+2

STEP 14

Apply the chain rule: dydt=dydu⋅dudt=10u9⋅(6t+2) \frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt} = 10u^9 \cdot (6t + 2) dtdy​=dudy​⋅dtdu​=10u9⋅(6t+2)

STEP 15

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Rewrite the limit using the factorized form: $\lim _{x \rightarrow 2}\left\{\frac{(x - 2)(x - 5)}{(x - 2)(x + 2)}\right\}$

Cancel out the common factor $(x - 2)$ : $\lim _{x \rightarrow 2}\left\{\frac{x - 5}{x + 2}\right\}$

Evaluate the second limit: ii) $\lim _{x \rightarrow 0}\left\{\frac{\sin 7 x}{x}\right\}$
We use the standard limit result $\lim _{u \rightarrow 0}\left\{\frac{\sin u}{u}\right\} = 1$ .

Rewrite the limit using the substitution $u = 7x$ : $\lim _{x \rightarrow 0}\left\{\frac{\sin 7 x}{x}\right\} = \lim _{x \rightarrow 0}\left\{\frac{\sin 7 x}{7x} \cdot 7\right\}$

Apply the standard limit result: $\lim _{x \rightarrow 0}\left\{\frac{\sin 7 x}{7x}\right\} = 1$
Thus, the second limit is: $\lim _{x \rightarrow 0}\left\{\frac{\sin 7 x}{x}\right\} = 7$

Differentiate $y=\left(3 t^{2}+2 t-9\right)^{10}$ using the chain rule.
Let $u = 3 t^{2} + 2 t - 9$ , then $y = u^{10}$ .

Differentiate $y$ with respect to $u$ : $\frac{dy}{du} = 10u^9$

Differentiate $u$ with respect to $t$ : $\frac{du}{dt} = 6t + 2$

Apply the chain rule: $\frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt} = 10u^9 \cdot (6t + 2)$