<s>[INST] Using power series to evaluate a limit If we want to evaluate the limit of $$\lim_{x\to 0} \frac{\ln(1-8x)+8x+32x^2}{3x^2}$$
I would imagine we need to use the power series expansion of the function $${\ln(1-8x)}$$
and then determine what cancels out.
The correct answer that I found using a calculator is $${-512/9}$$
 [/INST] $$\lim_{x\to 0} \frac{\ln(1-8x)+8x+32x^2}{3x^2}=\lim_{x\to 0} \frac{\ln(1-8x)}{3x^2}+\frac{8}{3x}+\frac{32}{3}$$
$$\ln(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\cdots-\frac{x^n}{n}$$
$$\ln(1-8x)=-8x-\frac{64x^2}{2}-\frac{512x^3}{3}-\cdots-\frac{8^nx^n}{n}$$
$$\frac{\ln(1-8x)}{3x^2}=-\frac{8}{3x}-\frac{32}{3}-\frac{512x}{9}-\cdots-\frac{8^nx^{n-2}}{3n}$$
$$\frac{\ln(1-8x)}{3x^2}+\frac{8}{3x}+\frac{32}{3}=-\frac{512x}{9}-\cdots-\frac{8^nx^{n-2}}{3n}$$
$$\lim_{x \to 0}\frac{\ln(1-8x)}{3x^2}+\frac{8}{3x}+\frac{32}{3}=\lim_{x \to 0}\left(-\frac{512x}{9}-\cdots-\frac{8^nx^{n-2}}{3n}\right)=0$$
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