# Use a graphing utility to graph the given function and

Use a graphing utility to graph the given function and the equations $$\displaystyle{y}={\left|{x}\right|}\ {\quad\text{and}\quad}\ {y}=-{\left|{x}\right|}$$ in the same viewing window. Using the graphs to observe the Squeeze Theorem visually, find $$\displaystyle\lim_{{{x}\Rightarrow{0}}}{f{{\left({x}\right)}}}$$.
$$\displaystyle{f{{\left({x}\right)}}}={\left|{x}\right|}{\cos{{x}}}$$

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maul124uk
Step 1
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
$$\displaystyle\lim_{{{x}\Rightarrow{0}}}{f{{\left({x}\right)}}}=\lim_{{{x}\Rightarrow{0}}}{x}{\cos{{\left({\frac{{{1}}}{{{x}}}}\right)}}}={0}$$
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censoratojk
Step 1
Given:
$$\displaystyle{f{{\left({x}\right)}}}={\left|{x}\right|}{\cos{{\left({x}\right)}}}$$
Use a graphing utility to graph the given function and the equations
$$\displaystyle{y}={\left|{x}\right|}$$
and $$\displaystyle{y}=-{\left|{x}\right|}$$
Step 2
Explanation:
The lower and upper functions have the same limit at $$\displaystyle{x}={0}$$.
The middle function has the same limit value because it is trapped between the two outer function.
Step 3
Squeez Theorem:
Suppose $$\displaystyle{f{{\left({x}\right)}}}\le{g{{\left({x}\right)}}}\le{h}{\left({x}\right)}$$ for all x in an open interval about "a".
$$\displaystyle\lim_{{{a}\Rightarrow{a}}}{f{{\left({x}\right)}}}=\lim_{{{x}\Rightarrow{a}}}{h}{\left({x}\right)}={L}$$
Then, $$\displaystyle\lim_{{{x}\Rightarrow{a}}}{g{{\left({x}\right)}}}={L}$$
At $$\displaystyle{x}={0},\lim_{{{x}\Rightarrow{0}}}{\left[-{\left|{x}\right|}\right]}=\lim_{{{x}\Rightarrow{0}}}{\left[{\left|{x}\right|}\right]}={0}$$
Then, $$\displaystyle\lim_{{{x}\Rightarrow{0}}}{\left|{x}\right|}{\cos{{\left({x}\right)}}}={0}$$
nick1337

$$\lim_{x \Rightarrow 0} f(x)=\lim_{x \Rightarrow 0} |x|$$
$$=0.1$$
$$\lim_{x \Rightarrow 0} f(x)=0$$