Use Taylor series to evaluate the following limits. lim_{xrightarrow4}frac{ln(x-3)}{x^2-16}

Question
Limits and continuity
asked 2020-11-22
Use Taylor series to evaluate the following limits.
\(\lim_{x\rightarrow4}\frac{\ln(x-3)}{x^2-16}\)

Answers (1)

2020-11-23
To evaluate the limit using Taylor series,
\(\lim_{x\rightarrow4}\frac{\ln(x-3)}{x^2-16}\)
\(\because\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...\)
\(\therefore\ln(x-3)=\ln(1+(x-4))\)
\(=(x-4)-\frac{(x-4)^2}{2}+\frac{(x-4)^3}{3}-\frac{(x-4)^4}{4}+...\)
\(\lim_{x\rightarrow4}\frac{\ln(x-4)}{x^2-16}=\frac{(x-4)-\frac{(x-4)^2}{2}+\frac{(x-4)^3}{3}-\frac{(x-4)^4}{4}+...}{(x-4)(x+4)}\)
\(=\frac{(x-4)[1-\frac{(x-4)}{2}+\frac{(x-4)^2}{3}-\frac{(x-4)^3}{4}+...]}{(x-4)(x+4)}\)
\(\frac{1-\frac{(x-4)}{2}+\frac{(x-4)^2}{3}-\frac{(x-4)^3}{4}+...}{x+4}\)
\(\frac{1-0+0-0+...}{4+4}\)
\(=\frac{1}{8}\)
\(\therefore\lim_{x\rightarrow4}\frac{\ln(x-3)}{x^2-16}=\frac{1}{8}\)
0

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