I'm trying to solve these two limits and I have been trying to use the L'Hospital rule but not really getting anywhere with it. Any help is appreciated.

(a) $\lim \limits_{x\to \infty} \dfrac {5x+n\ln(x)}{x+n^2\ln(x)}$

Using L'Hospitals rule I have:

$\lim \limits_{x\to \infty} \dfrac {5+\frac nx}{1+n^2/x}$

But I am not sure if this is correct

(b) $\lim \limits_{x\to \infty} \dfrac {x^n+x^2}{e^x+1}$

And from this I have:

$\lim \limits_{x\to \infty} \dfrac {nx^{n-1}+2x}{e^x}$

But again, I am not sure where to go.

  • $\begingroup$ Apologies, I did have that but mistyped it. $\endgroup$ – Aaron F Oct 27 '13 at 15:10
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    $\begingroup$ In the first one, what's keeping you from drawing conclusions regarding the limit? $\endgroup$ – Git Gud Oct 27 '13 at 15:14
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    $\begingroup$ Ah true! Could I conclude the answer to the first is 5? $\endgroup$ – Aaron F Oct 27 '13 at 15:15
  • $\begingroup$ You can. But if you're not sure, you should try to understand why. $\endgroup$ – Git Gud Oct 27 '13 at 15:16


In the first one divide the numerator & the denominator by $x$

$$\text{Now, }\lim_{x\to\infty}\frac{\ln x}x=\lim_{x\to\infty}\frac1x=0$$ as $\displaystyle \lim_{x\to\infty}\frac{\ln x}x$ is of the form $\frac\infty\infty$

In the second ,

if $n<0,\lim_{n\to\infty}x^n=0$ apply derivative twice

if $0<n\le 2,$ apply derivative twice

If $n>2$ apply derivative $\lceil n\rceil$ times


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