$\underset{x\to \mathrm{\infty}}{lim}(\frac{4}{5+\mathrm{cos}x})$ prove that the limit is not finite while $x\to \mathrm{\infty}$ using Cauchy definition

dglennuo
2022-05-28
Answered

$\underset{x\to \mathrm{\infty}}{lim}(\frac{4}{5+\mathrm{cos}x})$ prove that the limit is not finite while $x\to \mathrm{\infty}$ using Cauchy definition

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stacan6t

Answered 2022-05-29
Author has **13** answers

Suppose that limit L exists. For $\u03f5={\displaystyle \frac{1}{10}}$ such that $x>{M}_{1}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|f(x)-L|<{\displaystyle \frac{1}{10}}$

Let ${N}_{1}\in \mathbb{N}$ be such that $2\pi {N}_{1}>{M}_{1}$. Then if

$n>{N}_{1}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}2n\pi >2\pi {N}_{1}>{M}_{1}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|f(2n\pi )-L|<{\displaystyle \frac{1}{10}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|{\displaystyle \frac{2}{3}}-L|=|{\displaystyle \frac{4}{5+\mathrm{cos}(2n\pi )}}-L|<{\displaystyle \frac{1}{10}}$ Next for

$\u03f5={\displaystyle \frac{1}{15}},\mathrm{\exists}{M}_{2}$ such that $x>{M}_{2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|f(x)-L|<{\displaystyle \frac{1}{15}}$. Again, let ${N}_{2}\in \mathbb{N}$ be such that $(2{N}_{2}+1)\pi >{M}_{2}$. So if

$n>{N}_{2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}(2n+1)\pi >(2{N}_{2}+1)\pi >{M}_{2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|f((2n+1)\pi )-L|<{\displaystyle \frac{1}{15}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|1-L|<{\displaystyle \frac{1}{15}}$

Thus if $n>{N}_{1}+{N}_{2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\displaystyle \frac{1}{3}}\le |1-L|+|{\displaystyle \frac{2}{3}}-L|<{\displaystyle \frac{1}{10}}+{\displaystyle \frac{1}{15}}={\displaystyle \frac{1}{6}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\displaystyle \frac{1}{3}}<{\displaystyle \frac{1}{6}}$

Contradiction. So the limit Lcannot exists

Let ${N}_{1}\in \mathbb{N}$ be such that $2\pi {N}_{1}>{M}_{1}$. Then if

$n>{N}_{1}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}2n\pi >2\pi {N}_{1}>{M}_{1}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|f(2n\pi )-L|<{\displaystyle \frac{1}{10}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|{\displaystyle \frac{2}{3}}-L|=|{\displaystyle \frac{4}{5+\mathrm{cos}(2n\pi )}}-L|<{\displaystyle \frac{1}{10}}$ Next for

$\u03f5={\displaystyle \frac{1}{15}},\mathrm{\exists}{M}_{2}$ such that $x>{M}_{2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|f(x)-L|<{\displaystyle \frac{1}{15}}$. Again, let ${N}_{2}\in \mathbb{N}$ be such that $(2{N}_{2}+1)\pi >{M}_{2}$. So if

$n>{N}_{2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}(2n+1)\pi >(2{N}_{2}+1)\pi >{M}_{2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|f((2n+1)\pi )-L|<{\displaystyle \frac{1}{15}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}|1-L|<{\displaystyle \frac{1}{15}}$

Thus if $n>{N}_{1}+{N}_{2}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\displaystyle \frac{1}{3}}\le |1-L|+|{\displaystyle \frac{2}{3}}-L|<{\displaystyle \frac{1}{10}}+{\displaystyle \frac{1}{15}}={\displaystyle \frac{1}{6}}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{\displaystyle \frac{1}{3}}<{\displaystyle \frac{1}{6}}$

Contradiction. So the limit Lcannot exists

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I want to find the general antiderivative of $f(x)=x(6-x{)}^{2}$. However, I keep getting it wrong.

I am new to antiderivatives, but I think the first thing I should do is differentiate.

$\frac{d}{dx}(6-x{)}^{2}=2(6-x)\cdot -1=-2(6-x)$

${f}^{\prime}(x)=[(6-x{)}^{2}\cdot 1]+[-2(6-x)\cdot x]=(6-x{)}^{2}-2x(6-x)$

According to the antiderivative power rule, when $n\ne -1$, $\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C$.

So it seems like $\int (6-x{)}^{2}dx=\frac{(6-x{)}^{3}}{3}+C$.

However, I can't find a rule that seems like it would work with $-2x(6-x)$. The closest thing that I can find is the "Multiplication by Constant Rule", $\int cf(x)dx=c\cdot \int (f(x))dx$, but I'm not sure if I'm allowed to change $2x(6-x)$ into the form $2(6x-{x}^{2})$.

$\int 2(6x-{x}^{2})$

$=2\cdot \int (6x-{x}^{2})$

$=2(6\int x-\int {x}^{2})$

$=2(\frac{6{x}^{2}}{2}-\frac{{x}^{3}}{3})+C$

But $\frac{(6-x{)}^{3}}{3}-(6{x}^{2}-\frac{2{x}^{3}}{3})+C$ is incorrect.

I want to find the general antiderivative of $f(x)=x(6-x{)}^{2}$. However, I keep getting it wrong.

I am new to antiderivatives, but I think the first thing I should do is differentiate.

$\frac{d}{dx}(6-x{)}^{2}=2(6-x)\cdot -1=-2(6-x)$

${f}^{\prime}(x)=[(6-x{)}^{2}\cdot 1]+[-2(6-x)\cdot x]=(6-x{)}^{2}-2x(6-x)$

According to the antiderivative power rule, when $n\ne -1$, $\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C$.

So it seems like $\int (6-x{)}^{2}dx=\frac{(6-x{)}^{3}}{3}+C$.

However, I can't find a rule that seems like it would work with $-2x(6-x)$. The closest thing that I can find is the "Multiplication by Constant Rule", $\int cf(x)dx=c\cdot \int (f(x))dx$, but I'm not sure if I'm allowed to change $2x(6-x)$ into the form $2(6x-{x}^{2})$.

$\int 2(6x-{x}^{2})$

$=2\cdot \int (6x-{x}^{2})$

$=2(6\int x-\int {x}^{2})$

$=2(\frac{6{x}^{2}}{2}-\frac{{x}^{3}}{3})+C$

But $\frac{(6-x{)}^{3}}{3}-(6{x}^{2}-\frac{2{x}^{3}}{3})+C$ is incorrect.

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(d) Using a graphing utility, graph the function h,

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