Ok, this example requires you to use knowledge of natural logs and the exponential function in order to simplify the problem into a solvable form.

Since e^(ln(x)) = x, you can convert the function into the following form:

lim x -> 0+ e^(ln[(cosx)^[1/(x^2)]]) , which appears insanely messy, but it will work out.

since ln(x)^a = a*ln(x), this becomes the following:

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

The exponential function is always continuous, which means the limit can be moved inside:

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

take the derivative of the numerator and the denominator (l'hopital's rule):

(I removed the e^(lim x -> 0+ part to make it easier to understand)

(-sinx * 1/cosx)/(2x)

(-tanx/(2x))

Now apply the rule again

(-sec^2 (x) / 2)

sec(0) = 1/cos(0) = 1, so:

(-sec^2 (x) / 2) = -1/2

This means lim x-> 0+ [ln(cosx)/x^2] = -1/2

NOW, remember this is just the exponent of the exponential function, so the limit actually equals:

e^(-1/2) = 1/sqrt(e)

And this is the answer you get.

I'm sorry it was so complicated... haha

Let y = (cos x)^(1/x²)

Then ln(y) = 1/x² ln(cos x) = ln(cos x) / x²

So now we will find limit of ln(cos x) / x² as x approaches 0+

Once this is found, we can then convert it back.

lim[x→0+] ln(cos x) / x² = ln(1)/0 = 0/0 . . . . . Use l'Hopital rule

= lim[x→0+] (-sin x / cos x) / 2x

= lim[x→0+] (-tan x) / 2x = 0/0 . . . . . . . . Use l'Hopital rule again

= lim[x→0+] (-sec²x) / 2

= -1/2

ln(y) = -1/2

y = e^(-1/2)

y = 1/√e

Since y = (cos x)^(1/x²) and lim[x→0+] y = 1/√e, then

lim[x→0+] (cos x)^(1/x²) = 1/√e

Check: http://www.wolframalpha.com/input/?i=lim+%28cos%28...

look for patrickJMT on youtube, he has some great stuff on L'H. i figure a video would help u better then me trying to write out a sol for you. good luck