/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 Evaluate the following limits us... [FREE SOLUTION] | 91影视

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Chapter 10: Problem 20

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

Short Answer

Expert verified
Answer: The limit is 1.

Step by step solution

01

Find Taylor series for \(\ln x\) around \(x=1\)

Recall that the Taylor series for a function \(f(x)\) around \(x=a\) is given by: $$f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!} (x - a)^k$$ Now let's find the Taylor series for \(\ln x\) around \(x=1\). - \(f(x) = \ln x\) - \(f'(x) = \frac{1}{x}\) - \(f''(x) = -\frac{1}{x^2}\) - \(f'''(x) = \frac{2}{x^3}\) - \(f^{(k)}(x) = (-1)^{k-1}\frac{(k-1)!}{x^k}\) for \(k \geq 1\) Now, let's plug these derivatives and \(x = 1\) into the Taylor series formula to get the series for \(\ln x\) around \(x=1\). $$\ln x = \sum_{k=1}^{\infty} (-1)^{k-1}\frac{(x - 1)^k}{k}$$
02

Simplify limit expression with Taylor series

Now, we'll use the Taylor series we just found to simplify the limit expression. $$\lim _{x \rightarrow 1} \frac{x-1}{\ln x} = \lim _{x \rightarrow 1} \frac{x-1}{\sum_{k=1}^{\infty} (-1)^{k-1}\frac{(x - 1)^k}{k}}$$ Divide both numerator and denominator by \((x-1)\): $$\lim _{x \rightarrow 1} \frac{1}{\sum_{k=1}^{\infty} (-1)^{k-1}\frac{(x - 1)^{k-1}}{k}}$$
03

Evaluate the limit

As \(x \rightarrow 1\), the expression inside the limit becomes: $$\lim _{x \rightarrow 1} \frac{1}{\sum_{k=1}^{\infty} (-1)^{k-1}\frac{(x - 1)^{k-1}}{k}} = \frac{1}{\sum_{k=1}^{\infty} (-1)^{k-1}\frac{0^{k-1}}{k}}$$ $$= \frac{1}{\sum_{k=1}^{\infty} (-1)^{k-1}\frac{0^{k-1}}{k}}= \frac{1}{\sum_{k=1}^{\infty} (-1)^{k-1}\frac{\delta_{k,1}}{k}} $$ Where \(\delta_{k,1}\) is the Kronecker delta, which takes the value \(1\) if \(k=1\) and \(0\) otherwise. The only nonzero term in the sum is for \(k=1\), so the limit is just: $$\lim _{x \rightarrow 1} \frac{x-1}{\ln x} = \boxed{1}$$

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Most popular questions from this chapter

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.) $$\sin 0.2$$

Use an appropriate Taylor series to find the first four nonzero terms of an infinite series that is equal to the following numbers. $$\ln \left(\frac{3}{2}\right)$$

a. Use any analytical method to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. In most cases you do not need to use the definition of the Taylor series coefficients. b. If possible, determine the radius of convergence of the series. $$f(x)=\left(1+x^{2}\right)^{-2 / 3}$$

Determine whether the following statements are true and give an explanation or counterexample. a. The function \(f(x)=\sqrt{x}\) has a Taylor series centered at 0 . b. The function \(f(x)=\csc x\) has a Taylor series centered at \(\pi / 2\) c. If \(f\) has a Taylor series that converges only on \((-2,2),\) then \(f\left(x^{2}\right)\) has a Taylor series that also converges only on (-2,2) d. If \(p(x)\) is the Taylor series for \(f\) centered at \(0,\) then \(p(x-1)\) is the Taylor series for \(f\) centered at 1 e. The Taylor series for an even function about 0 has only even powers of \(x\)

The inverse hyperbolic sine is defined in several ways; among them are $$\sinh ^{-1} x=\ln (x+\sqrt{x^{2}+1})=\int_{0}^{x} \frac{d t}{\sqrt{1+t^{2}}}$$ Find the first four terms of the Taylor series for \(\sinh ^{-1} x\) using these two definitions (and be sure they agree).
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