calculus
posted by Lid on .
using the squeeze theorem, find the limit as x>0 of x*e^[8sin(1/x)]
what i did was:
1<=sin(1/x)<=1
8<=8*sin(1/x)<=8
e^(8)<=e^[8*sin(1/x)]<=e^(8)
x*e^(8)<=x*e^[8*sin(1/x)]<=x*e^(8)
lim x>0 [x*e^(8)] = 0
lim x>0 [x*e^(8)] = 0
hence, lim x>0 (x*e^[8*sin(1/x)]) = 0
is my solution right? the answer is correct, i'm just not sure about the procedures

correct