BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 3.3, Problem 42E

a.

To determine

Find thevalue of the given limit using given data.

Expert Solution

Answer to Problem 42E

The limx0sin5xx=5 .

Explanation of Solution

Given:

The given limitis limx0sin5xx .

Calculation:

  limx0sin5xx

Substitute the value of 5x=θ .

  x=θ5

If x=0 , then θ=0 .

Now,

  limx0sin5xx=limθ0sinθθ5=5limθ0sinθθ[limθ0sinθθ=1]=51=5

Hence limx0sin5xx=5 .

b.

To determine

Find the value of the given derivative using part(a).

Expert Solution

Answer to Problem 42E

  ddx(sin5x)=5cos5x

Explanation of Solution

Given:

The given expressionis ddx(sin5x) .

Calculation:

  f'(x)=limh0f(x+h)f(x)hf(x)=sin5xf(x+h)=sin(5x+5h)f'(x)=limh0sin(5x+5h)sin5xh

Use formula sinxsiny=2cos(x+y2)sin(xy2) .

  f'(x)=limh02cos(5x+5h+5x2)sin(5x+5h5x2)hf'(x)=limh02cos(10x+5h2)sin(5h2)hf'(x)=limh02cos(5x+5h2)sin(5h2)hf'(x)=limh02cos(5x+5h2)limh0sin(5h2)2h2

If h0 , then h20 .

  f'(x)=limh02cos(5x+5h2)12limh20sin(5h2)h2

From part (a).

  limx0sin5xx=5

Now,

  f'(x)=cos(5x+0)5f'(x)=5cos5x

Hence ddx(sin5x)=5cos5x .

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