# l'hopital for numerator and denominator separately

Is the following limit gives zero ? $$\lim_{n\to\infty}\frac{ln(n)}{n-ln(n)}$$

By substitution it gives $$\frac{\infty}{\infty-\infty}$$

I think we can not apply l'hopital directly, we can apply l'hopital only when we have $$\frac{\infty}{\infty}$$ or $$\frac{0}{0}$$

so I divided numerator and denominator by n $$\lim_{n\to\infty}\frac{ln(n)/n}{1-ln(n)/n}$$

by substitution it gives $$\frac{\infty/\infty}{-\infty/\infty}$$ I separated then the limit of the quotient :

$$\frac{\lim_{n\to\infty} (ln(n)/n)}{ \lim_{n\to\infty} (1-ln(n)/n)}$$

Applying l'Hopital in numerator, it gives $$\lim_{n\to\infty}(1/n)=0$$

for denominator , can we say that $$\lim_{n\to\infty} (1-ln(n)/n) =1- \lim_{n\to\infty}(1/n)=1$$ ? i.e. I applied l'hopital for the part of denominator that gives $$-\infty/\infty$$

hence $$\lim_{n\to\infty}\frac{ln(n)}{n-ln(n)} =0$$

-Is this right to separate the limit of numerator and denominator ? what are the cases we can not separate the limit for a quotient ? -And is this right to apply l'hopital for the numerator and denominator(or a part of it) separately ? Note: If I apply l'hopital for the whole denominator it gives $$\lim_{n\to\infty}(-1/n)=0$$ , and we got zero from applying l'hopital on the numerator ..So we have quotient of 2 limits and both give zero.. I didn't have idea how to complete the solution after that ... there is why I applied l'hopital for a part of denominator not all terms of it.

• What do you think $\infty - \infty$ evaluates to? – ÍgjøgnumMeg Mar 16 '16 at 19:21
• I think $$\infty-\infty$$ is an indeterminate form but $$\infty+\infty=\infty$$ – MCS Mar 16 '16 at 19:32

## 2 Answers

+1) You may look at it this way: $\frac{\ln n}{n-\ln n}=\frac{1}{\frac{n}{\ln n}-1}$ Now if $n$ goes to infinity, it is clear that $\frac{n}{\ln n}$ goes to infinity (one can also "prove" that if needed but I assume this is obvious), hence the whole fraction goes to zero.

Since $n - \ln(n)\to\infty$ as $n\to\infty$, your original form is an $\infty/\infty$ indeterminate form. Apply L'hopital right away.