Limit as $x\to 2$ of $\frac{\mathrm{cos}(\frac{\pi}{x})}{x-2}$

Janet Forbes
2022-07-08
Answered

Limit as $x\to 2$ of $\frac{\mathrm{cos}(\frac{\pi}{x})}{x-2}$

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asked 2022-06-04

I am prompted to solve the following limit

$\underset{x\mapsto 0}{lim}(\mathrm{cos}(x{)}^{\frac{1}{{x}^{2}}})$

I try to approach this problem by doing

$\underset{x\mapsto 0}{lim}(-1+(1+\mathrm{cos}(x){)}^{\frac{\mathrm{cos}(x)}{\mathrm{cos}(x){x}^{2}}})$

where the limit of $1+\mathrm{cos}(x{)}^{\frac{1}{\mathrm{cos}(x)}}$ is e. So I would have to calcualte the limit of $\frac{\mathrm{cos}(x)}{{x}^{2}}$. Is this approach correct?

$\underset{x\mapsto 0}{lim}(\mathrm{cos}(x{)}^{\frac{1}{{x}^{2}}})$

I try to approach this problem by doing

$\underset{x\mapsto 0}{lim}(-1+(1+\mathrm{cos}(x){)}^{\frac{\mathrm{cos}(x)}{\mathrm{cos}(x){x}^{2}}})$

where the limit of $1+\mathrm{cos}(x{)}^{\frac{1}{\mathrm{cos}(x)}}$ is e. So I would have to calcualte the limit of $\frac{\mathrm{cos}(x)}{{x}^{2}}$. Is this approach correct?

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As per the definition of limits if $\underset{x\to a}{lim}f(x)=L$, then

$\mathrm{\forall}\epsilon >0\text{}\mathrm{\exists}\delta 0\text{}s.t0|x-a|\delta \text{}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\text{}0|f(x)-L|\epsilon $

$\mathrm{\forall}\epsilon >0\text{}\mathrm{\exists}\delta 0\text{}s.t0|x-a|\delta \text{}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\text{}0|f(x)-L|\epsilon $

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Why does $\underset{x\to \pi}{lim}\frac{\mathrm{sin}(mx)}{\mathrm{sin}(nx)}={(-1)}^{m-n}\phantom{\rule{thickmathspace}{0ex}}\frac{m}{n}$, for positive naturals m and n?