Math / Calculus

QuestionThe Taylor polynomials we've shown you so far really do closely resemble the functions they come from, close to their centres.
In fact, a degree $n$ -Taylor polynomial for $f(x)$ (when $f(x)$ is one of the functions whose Maclaurin polynomials are on the list you should know) at $x=0$ will approximate the function $f(x)$ on the order of $x^{n}$ . More precisely, $\lim _{x \rightarrow a}\left(\frac{f(x)-T_{n}(x)}{x^{n}}\right)=0$
Notice that the above limit is a $\frac{0}{0}$ form. So the fact that the overall limit is zero means that the numerator $f(x)-T_{n}(x)$ goes to zero faster than the denominator $x^{n}$ .
Using this, evaluate the following limits: (a) $\lim _{x \rightarrow 0} \frac{\cos (x)-1+\frac{x^{2}}{2}}{9 x^{4}}=$ $\square$ (b) $\lim _{x \rightarrow 0} \frac{e^{3 x}-1-3 x-\frac{9 x^{2}}{2}}{x^{3}}=$ $\square$ (c) $\lim _{x \rightarrow 0} \frac{\sin (x)-x+\frac{x^{3}}{6}}{2 x^{4}}=$ $\square$ (d) $\lim _{x \rightarrow 0} \frac{e^{x^{5}} \log \left(1+x^{7}\right)+\cos \left(x^{99}\right)-\sin \left(x^{7}\right)-1}{x^{12}}=$ $\square$

Studdy Solution

STEP 1

What is this asking? We need to evaluate some limits involving the difference between a function and its Taylor polynomial, using the fact that this difference goes to zero faster than $x^n$ , where $n$ is the degree of the polynomial. Watch out! We need to use the correct Taylor polynomial for each function, centered at $x = 0$ .
Also, remember that the limit being zero tells us something about the relative speeds at which the numerator and denominator approach zero.

STEP 2

1. Cosine Limit
2. Exponential Limit
3. Sine Limit
4. Combined Limit

STEP 3

We're looking at $\lim_{x \rightarrow 0} \frac{\cos(x) - 1 + \frac{x^2}{2}}{9x^4}$ .
The numerator looks a lot like the Taylor polynomial for $\cos(x)$ around $x = 0$ .

STEP 4

Remember, the Taylor expansion for $\cos(x)$ is $1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + \dots$ The numerator here is the difference between $\cos(x)$ and its second-degree Taylor polynomial.

STEP 5

The next term in the Taylor expansion is $\frac{x^4}{24}$ .
This means the numerator behaves like $\frac{x^4}{24}$ as $x$ approaches zero.

STEP 6

So, our limit becomes $\lim_{x \rightarrow 0} \frac{\frac{x^4}{24}}{9x^4}$ .
We can simplify this by dividing both the numerator and denominator by $x^4$ , which gives us $\lim_{x \rightarrow 0} \frac{\frac{1}{24}}{9} = \frac{1}{24 \cdot 9} = \frac{1}{216}$ .

STEP 7

Now we have $\lim_{x \rightarrow 0} \frac{e^{3x} - 1 - 3x - \frac{9x^2}{2}}{x^3}$ .
The numerator looks like the Taylor expansion for $e^{3x}$ around $x = 0$ .

STEP 8

The Taylor expansion for $e^{3x}$ is $1 + 3x + \frac{(3x)^2}{2} + \frac{(3x)^3}{6} + \dots$ or $1 + 3x + \frac{9x^2}{2} + \frac{27x^3}{6} + \dots$

STEP 9

The numerator is the difference between $e^{3x}$ and its second-degree Taylor polynomial.
The next term is $\frac{27x^3}{6}$ .

STEP 10

Our limit becomes $\lim_{x \rightarrow 0} \frac{\frac{27x^3}{6}}{x^3}$ .
Dividing both the numerator and denominator by $x^3$ gives us $\frac{27}{6} = \frac{9}{2}$ .

STEP 11

We have $\lim_{x \rightarrow 0} \frac{\sin(x) - x + \frac{x^3}{6}}{2x^4}$ .
The numerator looks like the Taylor expansion for $\sin(x)$ around $x = 0$ .

STEP 12

The Taylor expansion for $\sin(x)$ is $x - \frac{x^3}{6} + \frac{x^5}{120} - \dots$ The numerator is the difference between $\sin(x)$ and its third-degree Taylor polynomial.

STEP 13

The next term in the Taylor expansion is $\frac{x^5}{120}$ .
So, our limit becomes $\lim_{x \rightarrow 0} \frac{\frac{x^5}{120}}{2x^4}$ .

STEP 14

Dividing both the numerator and denominator by $x^4$ gives us $\lim_{x \rightarrow 0} \frac{\frac{x}{120}}{2} = \frac{0}{240} = 0$ .

STEP 15

We're looking at $\lim_{x \rightarrow 0} \frac{e^{x^5} \log(1+x^7) + \cos(x^{99}) - \sin(x^7) - 1}{x^{12}}$ .

STEP 16

Using Taylor expansions, $e^{x^5} \approx 1 + x^5$ , $\log(1+x^7) \approx x^7$ , $\cos(x^{99}) \approx 1 - \frac{x^{198}}{2}$ , and $\sin(x^7) \approx x^7$ .

STEP 17

Substituting these into the numerator, we get $(1+x^5)(x^7) + (1 - \frac{x^{198}}{2}) - x^7 - 1 \approx x^7 + x^{12} + 1 - \frac{x^{198}}{2} - x^7 - 1 = x^{12} - \frac{x^{198}}{2}$ .

STEP 18

Our limit becomes $\lim_{x \rightarrow 0} \frac{x^{12} - \frac{x^{198}}{2}}{x^{12}}$ .
Dividing by $x^{12}$ gives us $\lim_{x \rightarrow 0} 1 - \frac{x^{186}}{2} = 1$ .

STEP 19

(a) $1/216$ (b) $9/2$ (c) $0$ (d) $1$

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Studdy Solution

STEP 1

STEP 2

1. Cosine Limit2. Exponential Limit3. Sine Limit4. Combined Limit

STEP 3

We're looking at lim⁡x→0cos⁡(x)−1+x229x4\lim_{x \rightarrow 0} \frac{\cos(x) - 1 + \frac{x^2}{2}}{9x^4}limx→0​9x4cos(x)−1+2x2​​.The numerator looks a lot like the Taylor polynomial for cos⁡(x)\cos(x)cos(x) around x=0x = 0x=0.

STEP 4

Remember, the Taylor expansion for cos⁡(x)\cos(x)cos(x) is 1−x22+x424−x6720+…1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + \dots1−2x2​+24x4​−720x6​+… The numerator here is the difference between cos⁡(x)\cos(x)cos(x) and its second-degree Taylor polynomial.

STEP 5

The next term in the Taylor expansion is x424\frac{x^4}{24}24x4​.This means the numerator behaves like x424\frac{x^4}{24}24x4​ as xxx approaches **zero**.

STEP 6

STEP 7

Now we have lim⁡x→0e3x−1−3x−9x22x3\lim_{x \rightarrow 0} \frac{e^{3x} - 1 - 3x - \frac{9x^2}{2}}{x^3}limx→0​x3e3x−1−3x−29x2​​.The numerator looks like the Taylor expansion for e3xe^{3x}e3x around x=0x = 0x=0.

STEP 8

The Taylor expansion for e3xe^{3x}e3x is 1+3x+(3x)22+(3x)36+…1 + 3x + \frac{(3x)^2}{2} + \frac{(3x)^3}{6} + \dots1+3x+2(3x)2​+6(3x)3​+… or 1+3x+9x22+27x36+…1 + 3x + \frac{9x^2}{2} + \frac{27x^3}{6} + \dots1+3x+29x2​+627x3​+…

STEP 9

The numerator is the difference between e3xe^{3x}e3x and its second-degree Taylor polynomial.The next term is 27x36\frac{27x^3}{6}627x3​.

STEP 10

Our limit becomes lim⁡x→027x36x3\lim_{x \rightarrow 0} \frac{\frac{27x^3}{6}}{x^3}limx→0​x3627x3​​.Dividing both the numerator and denominator by x3x^3x3 gives us 276=92\frac{27}{6} = \frac{9}{2}627​=29​.

STEP 11

We have lim⁡x→0sin⁡(x)−x+x362x4\lim_{x \rightarrow 0} \frac{\sin(x) - x + \frac{x^3}{6}}{2x^4}limx→0​2x4sin(x)−x+6x3​​.The numerator looks like the Taylor expansion for sin⁡(x)\sin(x)sin(x) around x=0x = 0x=0.

STEP 12

The Taylor expansion for sin⁡(x)\sin(x)sin(x) is x−x36+x5120−…x - \frac{x^3}{6} + \frac{x^5}{120} - \dotsx−6x3​+120x5​−… The numerator is the difference between sin⁡(x)\sin(x)sin(x) and its third-degree Taylor polynomial.

STEP 13

The next term in the Taylor expansion is x5120\frac{x^5}{120}120x5​.So, our limit becomes lim⁡x→0x51202x4\lim_{x \rightarrow 0} \frac{\frac{x^5}{120}}{2x^4}limx→0​2x4120x5​​.

STEP 14

Dividing both the numerator and denominator by x4x^4x4 gives us lim⁡x→0x1202=0240=0\lim_{x \rightarrow 0} \frac{\frac{x}{120}}{2} = \frac{0}{240} = 0limx→0​2120x​​=2400​=0.

STEP 15

We're looking at lim⁡x→0ex5log⁡(1+x7)+cos⁡(x99)−sin⁡(x7)−1x12\lim_{x \rightarrow 0} \frac{e^{x^5} \log(1+x^7) + \cos(x^{99}) - \sin(x^7) - 1}{x^{12}}limx→0​x12ex5log(1+x7)+cos(x99)−sin(x7)−1​.

STEP 16

Using Taylor expansions, ex5≈1+x5e^{x^5} \approx 1 + x^5ex5≈1+x5, log⁡(1+x7)≈x7\log(1+x^7) \approx x^7log(1+x7)≈x7, cos⁡(x99)≈1−x1982\cos(x^{99}) \approx 1 - \frac{x^{198}}{2}cos(x99)≈1−2x198​, and sin⁡(x7)≈x7\sin(x^7) \approx x^7sin(x7)≈x7.

STEP 17

STEP 18

Our limit becomes lim⁡x→0x12−x1982x12\lim_{x \rightarrow 0} \frac{x^{12} - \frac{x^{198}}{2}}{x^{12}}limx→0​x12x12−2x198​​.Dividing by x12x^{12}x12 gives us lim⁡x→01−x1862=1\lim_{x \rightarrow 0} 1 - \frac{x^{186}}{2} = 1limx→0​1−2x186​=1.

STEP 19

(a) 1/2161/2161/216 (b) 9/29/29/2 (c) 000 (d) 111

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1. Cosine Limit
2. Exponential Limit
3. Sine Limit
4. Combined Limit

We're looking at $\lim_{x \rightarrow 0} \frac{\cos(x) - 1 + \frac{x^2}{2}}{9x^4}$ .
The numerator looks a lot like the Taylor polynomial for $\cos(x)$ around $x = 0$ .

Remember, the Taylor expansion for $\cos(x)$ is $1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + \dots$ The numerator here is the difference between $\cos(x)$ and its second-degree Taylor polynomial.

The next term in the Taylor expansion is $\frac{x^4}{24}$ .
This means the numerator behaves like $\frac{x^4}{24}$ as $x$ approaches zero.

Now we have $\lim_{x \rightarrow 0} \frac{e^{3x} - 1 - 3x - \frac{9x^2}{2}}{x^3}$ .
The numerator looks like the Taylor expansion for $e^{3x}$ around $x = 0$ .

The Taylor expansion for $e^{3x}$ is $1 + 3x + \frac{(3x)^2}{2} + \frac{(3x)^3}{6} + \dots$ or $1 + 3x + \frac{9x^2}{2} + \frac{27x^3}{6} + \dots$

The numerator is the difference between $e^{3x}$ and its second-degree Taylor polynomial.
The next term is $\frac{27x^3}{6}$ .

Our limit becomes $\lim_{x \rightarrow 0} \frac{\frac{27x^3}{6}}{x^3}$ .
Dividing both the numerator and denominator by $x^3$ gives us $\frac{27}{6} = \frac{9}{2}$ .

We have $\lim_{x \rightarrow 0} \frac{\sin(x) - x + \frac{x^3}{6}}{2x^4}$ .
The numerator looks like the Taylor expansion for $\sin(x)$ around $x = 0$ .

The Taylor expansion for $\sin(x)$ is $x - \frac{x^3}{6} + \frac{x^5}{120} - \dots$ The numerator is the difference between $\sin(x)$ and its third-degree Taylor polynomial.

The next term in the Taylor expansion is $\frac{x^5}{120}$ .
So, our limit becomes $\lim_{x \rightarrow 0} \frac{\frac{x^5}{120}}{2x^4}$ .

Dividing both the numerator and denominator by $x^4$ gives us $\lim_{x \rightarrow 0} \frac{\frac{x}{120}}{2} = \frac{0}{240} = 0$ .

We're looking at $\lim_{x \rightarrow 0} \frac{e^{x^5} \log(1+x^7) + \cos(x^{99}) - \sin(x^7) - 1}{x^{12}}$ .

Using Taylor expansions, $e^{x^5} \approx 1 + x^5$ , $\log(1+x^7) \approx x^7$ , $\cos(x^{99}) \approx 1 - \frac{x^{198}}{2}$ , and $\sin(x^7) \approx x^7$ .

Our limit becomes $\lim_{x \rightarrow 0} \frac{x^{12} - \frac{x^{198}}{2}}{x^{12}}$ .
Dividing by $x^{12}$ gives us $\lim_{x \rightarrow 0} 1 - \frac{x^{186}}{2} = 1$ .

(a) $1/216$ (b) $9/2$ (c) $0$ (d) $1$