   Chapter 2.6, Problem 38E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Find the limit or show that it does not exist. lim x → ∞ sin 2 x x 2 + 1

To determine

To find: The value of limxsin2xx2+1.

Explanation

Theorem used 1: The Squeeze Theorem

“If g(x)f(x)h(x) when x is near a (except possibly at a) and limxag(x)=limxah(x)=L then limxaf(x)=L.”

Theorem used 2: If r>0 is a rational number, then limx1xr=0.

Calculation:

Obtain the limit of the function as x approaches infinity.

Consider the function f(x)=sin2xx2+1.

Since the range of sin2x lies between 0 and 1.

Consider  0sin2x1.

Divide the above inequality by x2+1,

0x2+1sin2xx2+11x2+1 0sin2xx2+11x2+1

Let g(x)=0 and h(x)=1x2+1.

Take the limit of upper and lower inequalities as x approaches infinity.

As x goes to infinity, g(x) goes to zero. That is,

limxg(x)=limx(0)=0

As x goes to infinity h(x) goes to zero

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