Question

Find the limit...

Answer 1

find the limit as x approaches +infinity

\[\frac{ x^{2} \ln x} { e^{x} }\]

Answer 2

L'Hôpital, hopefully?

Answer 3

yep. I solved it and the answer I got is 0... don't know if it's right or wrong...

Answer 4

Let's find out :)
This limit
\[\Large \lim_{x\rightarrow +\infty}\frac{x^2 \ln(x)}{e^x}\] is indeterminate of the form
\[\Large \frac{\infty}{\infty}\]
Applying L'Hôpital has us differentiate the numerator and denominator, and then taking the limit as x goes to positive infinity, we get...
\[\Large \lim_{x\rightarrow+\infty}\frac{x + 2x\ln(x)}{3^x}\]I believe...

Answer 5

Crap, typo, I meant
\[\Large \lim_{x\rightarrow+\infty}\frac{x + 2x\ln(x)}{e^x}\]

Answer 6

watch the language mother hecker!!

Answer 7

yep.. but still, it's infinity over infinity, so I applied the rule again and what I got was, and then got infinity over infinity again.. This was what I got that amde me conclude that the limit is 0.

\[\frac{ 2 }{ x(e ^{x}) }\]

if the numerator is constant, and the denominator is infinity, can I conclude that the limit is 0?

Answer 8

Yup :)

Answer 9

Nicely done ^_^

Answer 10

phew, thought that was wrong... I'm making improvements!!!!!!!