Use a graphing utility to graph the given function and

Joseph Krupa 2021-12-18 Answered
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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Expert Answer

maul124uk
Answered 2021-12-19 Author has 737 answers
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
Answered 2021-12-20 Author has 4897 answers
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, \(\displaystyle\lim_{{{x}\Rightarrow{0}}}{\left|{x}\right|}{\cos{{\left({x}\right)}}}={0}\)
0
nick1337
Answered 2021-12-28 Author has 9467 answers

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

0

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