Posted by **Lid** on Tuesday, October 27, 2015 at 5:18pm.

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

- calculus -
**Steve**, Tuesday, October 27, 2015 at 9:05pm
correct

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