Expert: Abe Mantell - 7/7/2009

Question
How would I find the Limit as x approaches 0 of ((sin(x))/x)^(1/(x^2))

Thank you

Answer
Let y=((sin(x))/x)^(1/(x^2)), now take the "ln" of both sides to get:
ln(y)=(1/x^2)ln(sin(x)/x)=ln(sin(x)/x)/x^2, now it is an indeterminate
form of the type "zero/zero" (since limit as x->0 of sin(x)/x = 1).
Now use L'Hopital's rule when taking the limit.

So, lim(x->0) ln(y)
= lim(x->0) {[x/sin(x)]*[(cos(x)/x-sin(x)/x^2)*x/sin(x)}/(2x)
= lim(x->0) (1/2)*x*(cos(x)/x-sin(x)/x^2)/sin(x)^2
= lim(x->0) (1/2)*(x*cos(x)-sin(x))/(x*sin(x)^2), (multiplying by x/x)

Now use L'Hopital's rule again on the right hand side, to get
= lim(x->0) -(1/2)x*sin(x)/(sin(x)^2+2*x*sin(x)*cos(x))
= lim(x->0) -(1/2)/[sin(x)/x+2cos(x)]
= -(1/2)[1+2] = -1/6

So, lim(x->0) ln(y)=1/6, which gives lim(x->0) y = e^(-1/6)

OK?

Abe