Show that the following limit is true:

Question

 

lim(x->0) x⁸ sin(1/x) = 0

 

I said we can't set it up lim(x->0)  0^8 and lim(x->0) sin(1/0) because that leaves us with something undefined

 

However, since -1 ≤ sin(1/x) ≤ 1, we know that ? ≤ x⁸ sin(1/x) ≤ ?

 

I don't know what to put for in for the question marks. I initially put 0 ≤ x8 sin(1/x) ≤ 0 but that's not correct. -1 ≤ x8 sin(1/x) ≤ 1 is also not correct. Help!

Answer 1

-1 ≤ sin(1/x) ≤ 1

 

So, -x⁸ ≤ x⁸sin(1/x) ≤ x⁸

 

lim_x→0 (-x⁸) = lim_x→0 (x⁸) = 0

 

So, by the Squeeze Theorem, lim_x→0 (x⁸sin(1/x)) = 0