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Let $f:\mathbb{R^*} - >\mathbb{R}, f(x) =\sin (\frac{1+x}{\sqrt x}) $. Prove that $\lim_{x \to 0^+} f(x) $ doesn't exist.
My solution : Since $f$ is continuous, we have
$\lim_{x \to 0^+} f(x)=f(\lim_{x\to 0^+} (\frac{1+x}{\sqrt x}))=lim_{x\to \infty} f(x) $
However, the last limit doesn't exist and hence the result.

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We have that

$$\frac{1+x}{\sqrt x} \to \infty$$

then the limit doesn't exist, to prove that we can consider

  • $x_n=\frac1{(2\pi n)^2}\to 0^+\implies \frac{1+x_n}{\sqrt x_n}\sim2\pi n\to \infty \implies \sin (\frac{1+x_n}{\sqrt x_n})\to 0$

  • $x_n=\frac1{(\pi/2+2\pi n)^2}\to 0^+\implies \frac{1+x_n}{\sqrt x_n}\sim \frac{\pi}2+2\pi n\to \infty \implies \sin (\frac{1+x_n}{\sqrt x_n})\to 1$

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  • $\begingroup$ Thank you! What about my proof? Is it all right or does it have any flaws? $\endgroup$ – user69503 Nov 25 '18 at 20:05
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    $\begingroup$ @user69503 No your way is not a proof, it is only a correct guess. For a proof we need to exhibit at least 2 subsequences with different limits. $\endgroup$ – gimusi Nov 25 '18 at 20:06
  • $\begingroup$ Why doesn't it work?. Didn't I correctly show that the given limit is the same as $\lim_{x\to \infty} f(x) $ which doesn't exist? If I prove that the latter is true, does my way work? $\endgroup$ – user69503 Nov 25 '18 at 20:14
  • $\begingroup$ @user69503 Essentially we are proving that $\lim_{y\to \infty} \sin y$ doesn't exist therefore if you are assuming that as a given your proof would be fine (wealso need to observe that $\frac{1+x}{\sqrt x}$ is continuous). But if you are not assuming that you need to explicitely show that. $\endgroup$ – gimusi Nov 25 '18 at 20:19
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    $\begingroup$ @user69503 Be carefull with that, your teacher could consider that as an error if the assumption is not explicitely given as a hint. $\endgroup$ – gimusi Nov 25 '18 at 20:27

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