Thread: L'Hospital

calculate:

$\displaystyle \lim_{x \to 0} (cos2x)^\frac{1}{x^2}$



Answer:
$\displaystyle e^-4$

$\displaystyle \lim_{x\to 0}(cos(2x))^{\frac{1}{x^{2}}}$

The limit is $\displaystyle e^{-2}$, not $\displaystyle e^{-4}$

There are many ways to go about it.

Let $\displaystyle t=2x$

Then you get:

$\displaystyle \lim_{t\to 0}\left[(cos(t))^{\frac{1}{t^{2}}}\right]^{4}$

Rewrite as:

$\displaystyle \left[e^{\lim_{t\to 0}\frac{ln(cos(t))}{t^{2}}}\right]^{4}$

The inside limit is a 'famous' one which equals -1/2, which easily derived using L'Hopital. You can do it without, but it's easier with the ol' Hospital rule

So, we have:

$\displaystyle \left[e^{-1/2}\right]^{4}=e^{-2}=\frac{1}{e^{2}}$

You wouldn't have to make the sub:

Start out with:

$\displaystyle e^{\lim_{x\to 0}\frac{ln(cos(2x))}{x^{2}}}$

Now, use L'Hopital and get:

$\displaystyle e^{-\lim_{x\to 0}\frac{sin(2x)}{xcos(2x)}}$

I hope you can read that. Sure is tiny. I have tried to make it bigger to no avail.

thanks for help