1. ## limit problem :(

hey everyone, i am new here and i need help as i have final today.
I have no problem with dealing with limits, but i am having troubles with this one.

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

first, I know that this problem should be solved using l'hopital's rule.

as i did it using it, i got to lim [ - cos(x^2) . 2x . 2x . 2x ] / e^x
x-->0
so my answer would be zero.

is this right? because my teacher said in class that the answer for this should be 2 and i wonder how she got 2.

2. Originally Posted by weskerq8
hey everyone, i am new here and i need help as i have final today.
I have no problem with dealing with limits, but i am having troubles with this one.

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

first, I know that this problem should be solved using l'hopital's rule.

as i did it using it, i got to lim [ - cos(x^2) . 2x . 2x . 2x ] / e^x
x-->0
so my answer would be zero.

is this right? because my teacher said in class that the answer for this should be 2 and i wonder how she got 2.

First Step

Differentiate Numerator
its = cos(x^2)* 2x

Doing so with denominator
its =e^x - 1
_________________________________________________
Second Step

Differentiate Numerator again
i

$\frac{d}{dx}(cos(x^2)*2x)= -sin(x^2)*2x * 2x + cos(x^2)*2
$
--Use chain Rule

Doing so with denominator again
its = e^x
Now pt the limit.Answer = 2

3. Thank you very much for the help