Why does $\underset{x\to {0}^{+}}{lim}\mathrm{s}\mathrm{g}\mathrm{n}(x{\mathrm{sin}}^{2}\left(\frac{1}{x}\right))$ not exist?

Agostarawz
2022-07-02
Answered

Why does $\underset{x\to {0}^{+}}{lim}\mathrm{s}\mathrm{g}\mathrm{n}(x{\mathrm{sin}}^{2}\left(\frac{1}{x}\right))$ not exist?

You can still ask an expert for help

Shawn Castaneda

Answered 2022-07-03
Author has **17** answers

The limit doe not exist, because $f({x}_{n})=0$ if $\phantom{\rule{thickmathspace}{0ex}}{x}_{n}={\displaystyle \frac{1}{n\pi}}$ and $f(x)=1$ if $\phantom{\rule{thickmathspace}{0ex}}{x}_{n}={\displaystyle \frac{1}{\frac{\pi}{2}+n\pi}}\phantom{\rule{1em}{0ex}}(n\in \mathbf{N})$, and both sequences converge to 0

asked 2022-06-13

How to appraoch solving this series?

I am given the following series and asked to solve it. $\sum _{n=1}^{\mathrm{\infty}}{\displaystyle \frac{(-2{)}^{n}}{(2n+1)!}}$ I recognize that this series is somewhat similar to the Taylor series for sinx which is $\sum _{n=0}^{\mathrm{\infty}}(-1{)}^{n}{\displaystyle \frac{{x}^{2n+1}}{(2n+1)!}}$.

However, I am not really able to relate these two series in order to solve them, especially since my series starts at 1 and once I rewrite the sinx series to match that, I am completely lost.

I am given the following series and asked to solve it. $\sum _{n=1}^{\mathrm{\infty}}{\displaystyle \frac{(-2{)}^{n}}{(2n+1)!}}$ I recognize that this series is somewhat similar to the Taylor series for sinx which is $\sum _{n=0}^{\mathrm{\infty}}(-1{)}^{n}{\displaystyle \frac{{x}^{2n+1}}{(2n+1)!}}$.

However, I am not really able to relate these two series in order to solve them, especially since my series starts at 1 and once I rewrite the sinx series to match that, I am completely lost.

asked 2022-05-22

What are the steps to solve this limit $\underset{x\to 0}{lim}\frac{\mathrm{cos}(x+\pi /2)}{x}$?

asked 2022-06-26

Get the limit from

$\underset{x\to \mathrm{\infty}}{lim}(x{\int}_{\frac{\pi}{2}}^{\mathrm{arctan}\left(x\right)}\mathrm{sin}\left({t}^{2}\right)dt\phantom{\rule{mediummathspace}{0ex}})$

$\underset{x\to \mathrm{\infty}}{lim}(x{\int}_{\frac{\pi}{2}}^{\mathrm{arctan}\left(x\right)}\mathrm{sin}\left({t}^{2}\right)dt\phantom{\rule{mediummathspace}{0ex}})$

asked 2022-07-27

the lower limit is 0 the upper limit is x on the first and thirdintegral. the second integral lower limit is 0 andupper limit is 2x.

$\int \mathrm{sin}\theta d\theta \int d\psi \int {r}^{2}\ufeffdr$

$\int \mathrm{sin}\theta d\theta \int d\psi \int {r}^{2}\ufeffdr$

asked 2022-07-09

We have the following limits:

$\underset{x\to \frac{\pi}{4}}{lim}{\displaystyle \frac{1-{\mathrm{tan}}^{2}x}{\mathrm{cos}2x}}$ and $\underset{\theta \to 1}{lim}={\displaystyle \frac{\theta -1+\mathrm{sin}({\theta}^{2}-1)}{{\theta}^{2}-1}}$

Using L'Hôpital I obtained that the second limit is $\theta ={\displaystyle \frac{3}{2}}$. For the first limit I was able to rewrite it to $\underset{x\to \frac{\pi}{4}}{lim}(1-{\mathrm{tan}}^{2}(x))\mathrm{sec}(2x)$, but I don't really know what to do with that.

$\underset{x\to \frac{\pi}{4}}{lim}{\displaystyle \frac{1-{\mathrm{tan}}^{2}x}{\mathrm{cos}2x}}$ and $\underset{\theta \to 1}{lim}={\displaystyle \frac{\theta -1+\mathrm{sin}({\theta}^{2}-1)}{{\theta}^{2}-1}}$

Using L'Hôpital I obtained that the second limit is $\theta ={\displaystyle \frac{3}{2}}$. For the first limit I was able to rewrite it to $\underset{x\to \frac{\pi}{4}}{lim}(1-{\mathrm{tan}}^{2}(x))\mathrm{sec}(2x)$, but I don't really know what to do with that.

asked 2022-01-24

Solving the following limit without LHospitals rule: $\underset{x\to 0}{lim}\frac{\mathrm{sin}({x}^{2}+2)-\mathrm{sin}(x+2)}{x}$

asked 2022-05-13

If $f(x)={x}^{2}$ and $g(x)=x\mathrm{sin}(x)+\mathrm{cos}(x)$ then i have to find number of points(x) such that f(x)=g(x)

I write h(x)=f(x)−g(x). So $h(x)={x}^{2}-x\mathrm{sin}(x)-\mathrm{cos}(x)$

${h}^{\prime}(x)=x(2-\mathrm{cos}(x))$. Since $2-\mathrm{cos}x$ is bounded so By taking limits as x goes to plus and minus infinity h′(x) goes to plus and minus infinity.which means h has root in between.

I write h(x)=f(x)−g(x). So $h(x)={x}^{2}-x\mathrm{sin}(x)-\mathrm{cos}(x)$

${h}^{\prime}(x)=x(2-\mathrm{cos}(x))$. Since $2-\mathrm{cos}x$ is bounded so By taking limits as x goes to plus and minus infinity h′(x) goes to plus and minus infinity.which means h has root in between.