Joseph Krupa

Answered

2021-12-18

Use a graphing utility to graph the given function and the equations $y=\left|x\right|\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}y=-\left|x\right|$ in the same viewing window. Using the graphs to observe the Squeeze Theorem visually, find $\underset{x\Rightarrow 0}{lim}f\left(x\right)$ .

$f\left(x\right)=\left|x\right|\mathrm{cos}x$

Answer & Explanation

maul124uk

Expert

2021-12-19Added 35 answers

Step 1

Given function is

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

we have to find$\underset{x\Rightarrow 0}{lim}f\left(x\right)$

Step 2

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

$h\left(x\right)=x\mathrm{cos}\frac{1}{x},{y}_{1}=\left|x\right|\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}{y}_{2}=-\left|x\right|$

Since$-1\le \mathrm{cos}\left(\frac{1}{x}\right)\le 1\text{}f{\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}all\text{}x\ne 0$

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

But$\underset{x\Rightarrow 0}{lim}{y}_{2}=\underset{x\Rightarrow 0}{lim}{y}_{1}=0$

Therefore squeeze theorem can be use to concude that

$\underset{x\Rightarrow 0}{lim}f\left(x\right)=\underset{x\Rightarrow 0}{lim}x\mathrm{cos}\left(\frac{1}{x}\right)=0$

Given function is

we have to find

Step 2

We know that cosine function is always -1 and is given function is always between

Since

it follows that

But

Therefore squeeze theorem can be use to concude that

censoratojk

Expert

2021-12-20Added 46 answers

Step 1

Given:

$f\left(x\right)=\left|x\right|\mathrm{cos}\left(x\right)$

Use a graphing utility to graph the given function and the equations

$y=\left|x\right|$

and$y=-\left|x\right|$

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".

$\underset{a\Rightarrow a}{lim}f\left(x\right)=\underset{x\Rightarrow a}{lim}h\left(x\right)=L$

Then,$\underset{x\Rightarrow a}{lim}g\left(x\right)=L$

At$x=0,\underset{x\Rightarrow 0}{lim}[-\left|x\right|]=\underset{x\Rightarrow 0}{lim}\left[\left|x\right|\right]=0$

Then,$\underset{x\Rightarrow 0}{lim}\left|x\right|\mathrm{cos}\left(x\right)=0$

Given:

Use a graphing utility to graph the given function and the equations

and

Step 2

Explanation:

The lower and upper functions have the same limit at

The middle function has the same limit value because it is trapped between the two outer function.

Step 3

Squeez Theorem:

Suppose

Then,

At

Then,

nick1337

Expert

2021-12-28Added 573 answers

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