 Joseph Krupa

2021-12-18

Use a graphing utility to graph the given function and the equations in the same viewing window. Using the graphs to observe the Squeeze Theorem visually, find $\underset{x⇒0}{lim}f\left(x\right)$.
$f\left(x\right)=|x|\mathrm{cos}x$ maul124uk

Expert

Step 1
Given function is
$h\left(x\right)=x\mathrm{cos}\frac{1}{x}$
we have to find $\underset{x⇒0}{lim}f\left(x\right)$
Step 2
We know that cosine function is always -1 and is given function is always between $-|x|$ and |x| which both go to zero as $x⇒0$.

Since
it follows that ${y}_{2}\le h\left(x\right)\le {y}_{1}$ for all $n\ne 0$
But $\underset{x⇒0}{lim}{y}_{2}=\underset{x⇒0}{lim}{y}_{1}=0$
Therefore squeeze theorem can be use to concude that
$\underset{x⇒0}{lim}f\left(x\right)=\underset{x⇒0}{lim}x\mathrm{cos}\left(\frac{1}{x}\right)=0$ censoratojk

Expert

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

Expert

$\underset{x⇒0}{lim}f\left(x\right)=\underset{x⇒0}{lim}|x|$
$=0.1$
$\underset{x⇒0}{lim}f\left(x\right)=0$

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