Question
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
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
Answers
Steve
correct