How to find the limit $\lim_{x\to\infty}\frac{\ln(x^2+x+5)}{\ln(x^8-x+3)}$?

Question

I don't know how to approach this problem:

$$\lim_{x\to\infty}\frac{\ln(x^2+x+5)}{\ln(x^8-x+3)}$$

I tried reducing it to the form $\lim (1+x_n)^{1/x_n}=1$, but that didn't work.

Answer 1

Factoring, we get:

$$x^2+x+5=x^2(1+1/x+5/x^2).$$

Using the above and log rules:

$$\ln(x^2+x+5)=\ln(x^2)+\ln(1+1/x+5/x^2)=2 \ln(x)+\ln(1+1/x+5/x^2).$$

Similarly:

$$\ln(x^8-x+3)=\ln(x^8)+\ln(1/x^7+3/x^8)=8 \ln(x)+\ln(1+1/x^7+3/x^8).$$

Can you take it from here?

Answer 2

Here's a step by step solution:

Since both, numerator and denominator of $$\frac{\ln(x^2+x+5)}{\ln(x^8-x+3)}$$ tend to $\infty$ when $x \rightarrow \infty$, we can apply L'Hôpital's rule

\begin{align*} \lim_{x\rightarrow\infty}\frac{\ln(x^2+x+5)}{\ln(x^8-x+3)}&=\lim_{x\rightarrow\infty}\frac{\frac{d}{dx}\ln(x^2+x+5)}{\frac{d}{dx}\ln(x^8-x+3)}\\ &=\lim_{x\rightarrow\infty}\frac{\frac{2x+1}{x^2+x+5}}{\frac{8x^7-1}{x^8-x+3}}\\ &=\lim_{x\rightarrow\infty}\frac{(2x+1)(x^8-x+3)}{(x^2+x+5)(8x^7-1)}\\ &=\lim_{x\rightarrow\infty}\frac{2x^9+x^8-2x^2+5x+3}{8x^9+8x^8+40x^7-x^2-x-5}\\ &=\lim_{x\rightarrow\infty}\frac{2+\frac{1}{x}-\frac{2}{x^7}+\frac{5}{x^8}+\frac{3}{x^9}}{8+\frac{8}{x}+\frac{40}{x^2}-\frac{1}{x^7}-\frac{1}{x^8}-\frac{5}{x^9}}\\ &=\frac{1}{4} \end{align*}

Note: Observe, that in the calculation above only the terms with the highest power of $x$ are relevant for the result.