AboutAbe Mantell Expertise Hello,
I am a college professor of mathematics and regularly teach all levels
from elementary mathematics through differential equations, and would
be happy to assist anyone with such questions!
Experience Over 15 years teaching at the college level.
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)
