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

Your solution is correct! You have correctly applied the squeeze theorem to find the limit of the function as x approaches 0.

The squeeze theorem states that if f(x) <= g(x) <= h(x) for all x in some interval around a certain point, except possibly at x = a, and the limits of f(x) and h(x) as x approaches a both equal L, then the limit of g(x) as x approaches a also equals L.

In your solution, you have bounded the function x*e^[8*sin(1/x)] between two functions, x*e^(-8) and x*e^(8), and found the limits of both of these functions as x approaches 0 to be 0. Therefore, by applying the squeeze theorem, you can conclude that the limit of the given function as x approaches 0 is also 0.

Well done on correctly using the squeeze theorem to find the limit!