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Chapter 10: Problem 14

Evaluate the following limits using Taylor series. $$\lim _{x \rightarrow \infty} x \sin \frac{1}{x}$$

Short Answer

Expert verified
Question: Evaluate the limit of x * sin(1/x) as x approaches infinity using the Taylor series. Answer: 1

Step by step solution

01

Taylor series of sin(1/x)

The Taylor series for the sine function, sin(u), centered at 0 is given by: $$\sin(u) = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}u^{2n+1}$$ We want the Taylor series of sin(1/x) around x=0, so we replace u with 1/x: $$\sin \left(\frac{1}{x}\right) =\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}\left(\frac{1}{x}\right)^{2n+1}$$
02

Multiply by x

Now we need to multiply the above series by x: $$x\sin \left(\frac{1}{x}\right) = x \left(\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}\left(\frac{1}{x}\right)^{2n+1}\right)$$ Simplifying this expression, we get: $$x\sin \left(\frac{1}{x}\right) =\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}\frac{1}{x^{2n}}$$
03

Find the limit as x approaches infinity

We want to find the limit as x approaches infinity for the expression above: $$\lim_{x \rightarrow \infty} \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}\frac{1}{x^{2n}}$$ Notice that the first term of the series is when n=0: $$\frac{(-1)^0}{(2(0)+1)!}\frac{1}{x^{2(0)}} = \frac{1}{1!} = 1$$ As x approaches infinity, the remaining terms of the series with n鈮�1 will approach 0 because the denominators will become larger and larger. Hence, only the first term will have a significant value. $$\lim_{x \rightarrow \infty} x\sin \left(\frac{1}{x}\right) = 1$$
04

Conclusion

Therefore, using the Taylor series, we can find that: $$\lim _{x \rightarrow \infty} x \sin \frac{1}{x} = 1$$

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Most popular questions from this chapter

Use properties of power series, substitution, and factoring of constants to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. Use the Taylor series. $$(1+x)^{-2}=1-2 x+3 x^{2}-4 x^{3}+\cdots, \text { for }-1 < x < 1$$ $$\frac{1}{\left(4+x^{2}\right)^{2}}$$

Use an appropriate Taylor series to find the first four nonzero terms of an infinite series that is equal to the following numbers. $$\tan ^{-1}\left(\frac{1}{2}\right)$$

By comparing the first four terms, show that the Maclaurin series for \(\sin ^{2} x\) can be found (a) by squaring the Maclaurin series for \(\sin x,\) (b) by using the identity \(\sin ^{2} x=(1-\cos 2 x) / 2,\) or \((\mathrm{c})\) by computing the coefficients using the definition.

What is the minimum order of the Taylor polynomial required to approximate the following quantities with an absolute error no greater than \(10^{-3}\) ? (The answer depends on your choice of a center.) $$\sin 0.2$$

Errors in approximations Suppose you approximate \(\sin x\) at the points \(x=-0.2,-0.1,0.0,0.1,\) and 0.2 using the Taylor polynomials \(p_{3}=x-x^{3} / 6\) and \(p_{5}=x-x^{3} / 6+x^{5} / 120 .\) Assume that the exact value of \(\sin x\) is given by a calculator. a. Complete the table showing the absolute errors in the approximations at each point. Show two significant digits. $$\begin{array}{|c|l|l|} \hline x & \text { Error }=\left|\sin x-p_{3}(x)\right| & \text { Error }=\left|\sin x-p_{5}(x)\right| \\ \hline-0.2 & & \\ \hline-0.1 & & \\ \hline 0.0 & & \\ \hline 0.1 & & \\ \hline 0.2 & & \\ \hline \end{array}$$ b. In each error column, how do the errors vary with \(x\) ? For what values of \(x\) are the errors the largest and smallest in magnitude?
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