Problem 20 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Answer: The limit is 1.

Step 1: 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 \(x=1\). $$\ln x = \sum_{k=1}^{\infty} (-1)^{k-1}\frac{(x - 1)^k}{k}$$

Step 2: 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 \((x-1)\): $$\lim _{x \rightarrow 1} \frac{1}{\sum_{k=1}^{\infty} (-1)^{k-1}\frac{(x - 1)^{k-1}}{k}}$$

Step 3: 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: $$\lim _{x \rightarrow 1} \frac{x-1}{\ln x} = \boxed{1}$$