1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Apologies, I did have that but mistyped it. $\endgroup$ – Aaron F Oct 27 '13 at 15:10
  • 1
    $\begingroup$ In the first one, what's keeping you from drawing conclusions regarding the limit? $\endgroup$ – Git Gud Oct 27 '13 at 15:14
  • 1
    $\begingroup$ 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
1
$\begingroup$

HINT:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.