$\lim_{n \to \infty}\left(\frac {\ln(n^2+n+100)}{\ln(n^{100}+999n-1)}\right)$ I'm having trouble with the following limit: $$\lim_{n \to \infty}\left(\frac {\ln(n^2+n+100)}{\ln(n^{100}+999n-1)}\right)$$
I'd be grateful for any help. I've tried to write that as $$\lim_{n\to \infty}\left(\frac {\ln(n^2) + \ln(1+\frac 1 n+\frac {100}{n^2})}{\ln(n^{100})+\ln(1+\frac {999n}{n^{100}}-\frac{1}{n^{100}})}\right)$$ Now we know that two of those limits are equal to $0$, so that limit should be equal to $\lim_{n\to \infty}(\log_{n^{100}}n^2)= \frac{1}{50}$ ; But my question is: am I not making some mistakes here operating partly on limits and partly on values of those limits?
Thank you!
 A: You can make your proof rigorous, dividing both numerator and denominator by $\ln n$:
$$
\frac {\ln(n^2) + \ln(1+\frac 1 n+\frac {100}{n^2})}{\ln(n^{100})+\ln(1+\frac {999n}{n^{100}}-\frac{1}{n^{100}})} = \frac {2\ln(n) + \ln(1+\frac 1 n+\frac {100}{n^2})}{100\ln(n)+\ln(1+\frac {999n}{n^{100}}-\frac{1}{n^{100}})} =
$$
$$
= \frac {2 + \frac{\ln(1+\frac 1 n+\frac {100}{n^2})}{\ln n}}{100+\frac{\ln(1+\frac {999n}{n^{100}}-\frac{1}{n^{100}})}{\ln n}}.
$$
Now the limits of of numerator and denominator exist and equal to $2$ and $100$ respectively.
A: Use L'Hospital with
$$\lim_{x\to\infty}\frac{\log(x^2+x+100)}{\log(x^{100}+999x-1)}\stackrel{L'H}=\lim_{x\to\infty}\frac{2x+1}{x^2+x+100}\cdot\frac{x^{100}+999x-1}{100x^{99}+999}=\frac1{50}$$
Hint: remember that in limits as the above one, the degree of the numerator and denominator polynomials rule.
Your "trick" is fine...as long as you can justify it.
A: Your way is correct indeed from here
$$\frac {\ln(n^2+n+100)}{\ln(n^{100}+999n-1)}=\frac {\ln n^2+\ln\left(1+\frac1n+\frac{100}{n^2}\right)}{\ln n^{100}+\ln\left(1+\frac{999}n-\frac{1}{n^2}\right)}=\dots$$
we have
$$\dots=\frac {2\ln n+\ln\left(1+\frac1n+\frac{100}{n^2}\right)}{100 \ln n+\ln\left(1+\frac{999}n-\frac{1}{n^2}\right)} \to \frac1{50}$$
