# Math Help - limits

1. ## limits

can somebody please explain this problem to me:

lim [ (x(e^(-2x+1))) / ((x^2)+x)]
x-->0

i get 0/0 = undefined
but my teacher tells me that it is not correct.

thank you

2. Nevermind.

3. Originally Posted by rahul14
lim [ (x(e^(-2x+1))) / ((x^2)+x)]
x-->0
$\lim _{x \to 0} \frac{xe^{-2x+1}}{x^2+x}=\lim _{x \to 0}\frac{e^{-2x+1}}{x+1}\to e$

4. Originally Posted by rahul14
can somebody please explain this problem to me:

lim [ (x(e^(-2x+1))) / ((x^2)+x)]
x-->0

i get 0/0 = undefined
but my teacher tells me that it is not correct.

thank you
do you mean
$\lim_{x \to 0} \frac {x.e^{-2x+1}}{x^2+x}$

if yes , then
$\lim_{x \to 0} \frac {x.e^{-2x+1}}{x^2+x}=\lim_{x \to 0} \frac {x.e^{-2x+1}}{x(x+1)}$

$= \lim_{x \to 0} \frac {e^{-2x+1}}{x+1}= \frac {e^{0+1}}{0+1} =e$

5. Originally Posted by Defunkt
Yes, $\frac{0}{0}$ is correct. Have you learned l'Hôpital's rule?
No, 0/0 is NOT correct for the limit! It's not even a number. And I surely would not recommend "L'Hopital's rule" for something as simple as this.

$\frac{xe^{-2x+1}}{x^2+ x}= \frac{xe^{-2x+1}}{x(x+1)}$
which is equal to $\frac{e^{-2x+1}}{x+1}$ as long as x is not 0. They have the same limit so just put x= 0 in $\frac{e^{-2x+1}}{x+1}$