Question

I don't get trigonometric limits..

1. Use the squeeze theorem to evaluate the limit as x approaches 0 of x^2sin(1/x)

2. limit as h approaches 0 of [sin(5h)]/3h

3. limit as x approaches 0 of (x^2)/sin^2(x)

4. limit as t approaches 0 of [cos(t)-cos^2(t)]/t

5. Use the squeeze theorem to evaluate the limit as x approaches 1 of (x-1)sin(pi/x-1)

Thank you!

Answer 1

one Question at one post
1) the greatest value of sin(something) is 1 and the smallest value is -1. squeeze sin(something) between -1 and 1

Answer 2

\[ -1 \leq \sin(1/x) \leq +1 \\
- x^2\leq x^2\sin(1/x) \leq x^2 \]
then put limits on both sides
\[ \lim_{x->0}
- x^2\leq \lim_{x->0}x^2\sin(1/x) \leq \lim_{x->0}x^2
\\

0\leq \lim_{x->0}x^2\sin(1/x) \leq 0\]