Question

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

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

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