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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!

user
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EngineerInProgress
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  • I was wondering why putting the limits inside the log function is wrong? $$\lim_{n\to \infty}\left(\frac {\ln(n^2)+\ln(1+\frac 1 n+\frac {100}{n^2})}{\ln(n^{100})+\ln(1+\frac {999n}{n^{100}}-\frac{1}{n^{100}})}\right)\color{red}{=}\left(\frac {\ln(\lim_{n\to \infty}n^2) + \ln(\lim_{n\to \infty}1+\lim_{n\to \infty}\frac {1}{n}+\lim_{n\to \infty}\frac {100}{n^2})}{\ln(\lim_{n\to \infty}n^{100})+\ln(\lim_{n\to \infty}1+\lim_{n\to \infty}\frac {999n}{n^{100}}-\lim_{n\to \infty}\frac{1}{n^{100}})}\right).$$ Maybe someone can give me a rigorous explanation for this? – Philipp Oct 19 '19 at 11:37

3 Answers3

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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.

Virtuoz
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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.

DonAntonio
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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

$$\dots=\frac {2\ln n+\ln\left(1+\frac1n+\frac{100}{n^2}\right)}{100 \ln n+\ln\left(1+\frac{999}n-\frac{1}{n^2}\right)} \to \frac1{50}$$

user
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