Given function is

\(\displaystyle{h}{\left({x}\right)}={x}{\cos{{\frac{{{1}}}{{{x}}}}}}\)

we have to find \(\displaystyle\lim_{{{x}\Rightarrow{0}}}{f{{\left({x}\right)}}}\)

Step 2

We know that cosine function is always -1 and is given function is always between \(\displaystyle-{\left|{x}\right|}\) and |x| which both go to zero as \(\displaystyle{x}\Rightarrow{0}\).

\(\displaystyle{h}{\left({x}\right)}={x}{\cos{{\frac{{{1}}}{{{x}}}}}},{y}_{{{1}}}={\left|{x}\right|}\ {\quad\text{and}\quad}\ {y}_{{{2}}}=-{\left|{x}\right|}\)

Since \(\displaystyle-{1}\le{\cos{{\left({\frac{{{1}}}{{{x}}}}\right)}}}\le{1}\ {f}{\quad\text{or}\quad}\ {a}{l}{l}\ {x}\ne{0}\)

it follows that \(\displaystyle{y}_{{{2}}}\le{h}{\left({x}\right)}\le{y}_{{{1}}}\) for all \(\displaystyle{n}\ne{0}\)

But \(\displaystyle\lim_{{{x}\Rightarrow{0}}}{y}_{{{2}}}=\lim_{{{x}\Rightarrow{0}}}{y}_{{{1}}}={0}\)

Therefore squeeze theorem can be use to concude that

\(\displaystyle\lim_{{{x}\Rightarrow{0}}}{f{{\left({x}\right)}}}=\lim_{{{x}\Rightarrow{0}}}{x}{\cos{{\left({\frac{{{1}}}{{{x}}}}\right)}}}={0}\)