use the squeez theorem to find lim x0 xsin1x HINT 1 less tha

use the squeez theorem to find lim x0 (x.sin(1/x))
HINT# -1 less than equal to sin(x) less than equal to 1 or    -1<= sin(1) <=1          hope you got me ......

Solution

start with -1 sin t 1 for all t (or its cosine counterpart).

Let t = 1/x:
-1 sin(1/x) 1 for all x 0.

Multiply through by x^2:
-x^2 x^2 sin(1/x) x^2 for all x 0.

Since lim(x0) ± x^2 = 0, we conclude by the Squeeze Theorem that
lim(x0) x^2 sin(1/x) = 0.