Answer
$\displaystyle\lim_{h\rightarrow 0} \sin h\cos\dfrac{1}{h}=0$
Work Step by Step
We have to estimate the limit:
$\displaystyle\lim_{h\rightarrow 0} \sin h\cos\dfrac{1}{h}$
Compute $\sin h\cos\dfrac{1}{h}$ for values of $h$ close to 0:
$\sin (-0.01)\cos\dfrac{1}{-0.01}\approx -0.00862304$
$\sin (-0.001)\cos\dfrac{1}{-0.001}\approx -0.00056238$
$\sin (-0.0001)\cos\dfrac{1}{-0.0001}\approx 0.00009522$
$\sin (-0.00001)\cos\dfrac{1}{-0.00001}\approx 0.00000999$
$\sin (0.00001)\cos\dfrac{1}{0.00001}\approx -0.00000999$
$\sin (0.0001)\cos\dfrac{1}{0.0001}\approx -0.00009522$
$\sin (0.001)\cos\dfrac{1}{0.001}\approx 0.00056238$
$\sin (0.01)\cos\dfrac{1}{0.01}\approx 0.00862304$
Therefore we got:
$\displaystyle\lim_{h\rightarrow 0} \sin h\cos\dfrac{1}{h}=0$
