# Math Help - L Hopital Rule

1. ## L Hopital Rule

Ok I have this

as lim x --> 0 for

$\frac{sin^2x}{x^2}$

and $sin^2(0)=0$

and $0^2=0$

so the derivative as lim x-->0

$\frac{2sinxcosx}{2x}$

and i plug in 0 i get 0 so the limit is 0?

2. Originally Posted by sk8erboyla2004
...
so the derivative as lim x-->0

$\frac{2sinxcosx}{2x}$

and i plug in 0 i get 0 so the limit is 0?
No, you get $\frac00,$ which is an indeterminate form.

$\frac{\sin^2x}{x^2}=\left(\frac{\sin x}x\right)^2.$

There is a special trigonometric limit that you can apply here.

3. Ok so where I can from here?

4. $\lim_{x\to0}\frac{\sin^2x}{x^2}=\lim_{x\to0}\left( \frac{\sin x}x\right)^2$

$=\left(\lim_{x\to0}\frac{\sin x}x\right)^2$

$=1^2=1$

5. So do you only apply the L Hopital Rule When the lim of each denom and num are not both 0 or postive or negative infinity ?

how about for this one ?

lim x-->0

$\frac{e^{-x}}{x}$

I do not see how you got 1 it is my fault because we are taught to take the limit of each num and denom

6. Originally Posted by sk8erboyla2004
So do you only apply the L Hopital Rule When the lim of each denom and num are not both 0 or postive or negative infinity ?
L'Hôpital's rule is not needed for this limit. You can use it (twice), but that is more work. You should have been taught the following two special trigonometric limits.

$\lim_{x\to0}\frac{\sin x}x=1$

$\lim_{x\to0}\frac{1-\cos x}x=0$

how about for this one ?

lim x-->0

$\frac{e^{-x}}{x}$
Rewrite, $\frac{e^{-x}}x=\frac1{xe^x}.$ Since $e^x$ is always positive, it should be clear that as $x\to0,\;\frac1{xe^x}$ tends to $\infty$ from the right, and $-\infty$ from the left. This limit does not exist.

7. I was never taught the special trig limits for the first problem I did go on step further and got 1

because i noticed L Hoptial rule can no longer be used until you reach a constant in either denom or numerator

8. Originally Posted by sk8erboyla2004
I was never taught the special trig limits for the first problem I did go on step further and got 1

because i noticed L Hoptial rule can no longer be used until you reach a constant in either denom or numerator
If you must use l'Hopital's Rule, you should note that

$\frac{\sin^2 x}{x^2} \longrightarrow \frac{2 \sin x \cos x}{2x} = \frac{\sin (2x)}{2x} \longrightarrow \frac{2 \cos (2x)}{2} = \cos (2x)$

9. I think I get the geist of the rule

when ever you have f(x)/g(x) and both limits are 0/0, inf/inf, and -inf/-inf

then you apply the rule