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Chapter 10: Problem 9

Evaluate the following limits using Taylor series. $$\lim _{x \rightarrow 0} \frac{-x-\ln (1-x)}{x^{2}}$$

Short Answer

Expert verified
Answer: The limit is equal to \(- \frac{1}{2}\).

Step by step solution

01

Recall Taylor series expansion for the natural logarithm

Recall the Taylor series expansion for \(\ln(1+x)\) around zero is given by $$\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}x^n}{n} = x-\frac{x^2}{2}+\frac{x^3}{3} -\frac{x^4}{4}+...$$ Now, we have \(\ln(1-x)\) instead of \(\ln(1+x)\). So we will replace \(x\) by \(-x\) in the above series and simplify.
02

Taylor series expansion for \(\ln(1-x)\)

Replace \(x\) by \(-x\) in the expansion for \(\ln(1+x)\). $$\ln(1-x) = (-x)-\frac{(-x)^2}{2}+\frac{(-x)^3}{3} -\frac{(-x)^4}{4}+... = -x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}+...$$
03

Substitute the expansion obtained in Step 2 for \(\ln(1-x)\) in the limit

Replace the \(\ln(1-x)\) in the original limit expression with the Taylor series expansion obtained in Step 2: $$\lim _{x \rightarrow 0} \frac{-x-\ln (1-x)}{x^{2}} = \lim _{x \rightarrow 0} \frac{-x-\left(-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}+...\right)}{x^{2}}$$ Simplify the expression inside the limit: $$\lim _{x \rightarrow 0} \frac{-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}+...}{x^{2}}$$
04

Factor out the common factor \(x^2\) from each term in the numerator

Factor out \(x^2\) from each term in the numerator: $$\lim _{x \rightarrow 0} \frac{x^2\left(-\frac{1}{2}-\frac{x}{3}-\frac{x^2}{4}+...\right)}{x^{2}}$$ Now, cancel out \(x^2\) from numerator and denominator: $$\lim _{x \rightarrow 0} \left(-\frac{1}{2}-\frac{x}{3}-\frac{x^2}{4}+...\right)$$
05

Evaluate the limit

Now, as \(x\) approaches zero, all terms involving \(x\) will go to zero. Therefore, the limit will be: $$\lim _{x \rightarrow 0} \left(-\frac{1}{2}-\frac{x}{3}-\frac{x^2}{4}+...\right) = -\frac{1}{2}$$ So, the limit \(\lim _{x \rightarrow 0} \frac{-x-\ln (1-x)}{x^{2}}\) is equal to \(- \frac{1}{2}\).

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