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Chapter 4: Problem 89

Use analytical methods to evaluate the following limits. $$\lim _{x \rightarrow \infty} x^{3}\left(\frac{1}{x}-\sin \frac{1}{x}\right)$$

Short Answer

Expert verified
Answer: \(\frac{1}{6}\)

Step by step solution

01

Note the behavior of the first factor x^3 as x approaches infinity

The first factor in the function is \(x^3\). As x approaches infinity, \(x^3\) will also approach infinity.
02

Consider the Taylor series expansion of sin(x) to help with the second factor

To simplify the second factor, we can use the first few terms of the Taylor series expansion for sin(x) around x = 0. The Taylor series for sin(x) is: $$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$$ Now, replace x with 1/x in the Taylor series: $$\sin \frac{1}{x} = \frac{1}{x} - \frac{1}{3!x^3} + \frac{1}{5!x^5} - \dots$$
03

Subtract sin(1/x) from 1/x in the second factor using Taylor series expansion

Now, subtract the Taylor series of \(\sin\frac{1}{x}\) from \(\frac{1}{x}\): $$\frac{1}{x} - \sin \frac{1}{x} = \frac{1}{x} - \left(\frac{1}{x} - \frac{1}{3!x^3} + \frac{1}{5!x^5}\right)$$ This simplifies to: $$\frac{1}{x} - \sin \frac{1}{x} = \frac{1}{3!x^3} - \frac{1}{5!x^5} + \dots$$
04

Multiply x^3 with the simplified expression obtained in Step 3

Now, we will multiply \(x^3\left(\frac{1}{x}-\sin\frac{1}{x}\right)\) $$x^3\left(\frac{1}{3!x^3} - \frac{1}{5!x^5} + \dots\right) = \frac{x^3}{3!x^3} - \frac{x^3}{5!x^5} + \dots$$ This simplifies to: $$\frac{1}{3!} - \frac{1}{5!x^2} + \dots$$
05

Evaluate the limit as x approaches infinity

Now, we can easily evaluate the limit as x approaches infinity: $$\lim _{x \rightarrow \infty} x^{3}\left(\frac{1}{x}-\sin\frac{1}{x}\right) = \lim _{x \rightarrow \infty} \left(\frac{1}{3!} - \frac{1}{5!x^2} + \dots\right) = \frac{1}{3!}$$ The limit of the given function when x approaches infinity is \(\boxed{\frac{1}{6}}\).

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