$\lim_{n \to \infty}\left(\frac {\ln(n^2+n+100)}{\ln(n^{100}+999n-1)}\right)$ 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!

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.
