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I used $u$ substitution for the limit $$\lim_{x \to\infty}x^2 \sin\frac{1}{x}$$ and got the limit does not exist by saying $u=\frac{1}{x}$. Is this correct and if so would that mean $\lim\limits_{x \to\infty}x^3 \sin\frac{1}{x^2}$ and $\lim\limits_{x \to\infty}x^3 \sin\frac{1}{x}$ also don't exist?

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  • $\begingroup$ How did you do your substitution? $\endgroup$ – Frank Vel Jan 20 '15 at 22:32
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    $\begingroup$ It is correct that the limits don't exist (or to say they are all $+\infty$ which is one way for a limit not to exist), but "by saying $u=\frac1x$" is a bit too terse to work as an argument. At best you'll end up needing to show that $\frac1{u^2}\sin u$ has no limit for $u\to 0^+$, and that deserves some words of its own. $\endgroup$ – Henning Makholm Jan 20 '15 at 22:33
  • $\begingroup$ Using \sin instead of sin results in nicer formatting. This works for various common functions. $\endgroup$ – quid Jan 20 '15 at 22:38
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Write the limit with the substitution $t=1/x$: $$ \lim_{t\to0^+}\frac{\sin t}{t^2}=\lim_{t\to0^+}\frac{\sin t}{t}\frac{1}{t} $$ The first factor has limit $1$, so…

The same substitution will make $$ \lim_{x\to\infty}x^3\sin\frac{1}{x}= \lim_{t\to0^+}\frac{\sin t}{t}\frac{1}{t^2} $$ so…

For the case of $\lim_{x\to\infty}x^3\sin\frac{1}{x^2}$, the substitution would give $$ \lim_{x\to\infty}x^3\sin\frac{1}{x^2}= \lim_{t\to0^+}\frac{\sin t^2}{t^2}\frac{1}{t} $$ and we'd be back to the first case.

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Since $\sin\frac{1}{x}\to0$ and $x^2\to\infty$, we use l'Hôpital's theorem: $$\lim_{x\to\infty}\frac{\sin\frac{1}{x}}{\frac{1}{x^2}}=\lim_{x\to\infty}\frac{-\frac{1}{x^2}\cos\frac{1}{x}}{-\frac{2}{x^3}}=\lim_{x\to\infty}\frac{x}{2}\cos\frac{1}{x}=\infty$$ since $\cos\frac{1}{x}\to1$ and $\frac{x}{2}\to\infty$, as $x \rightarrow \infty.$

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Suppose to the contrary that $\lim_{x\to \infty} x^2\sin(1/x)$ exists and equals $L$. Since $\lim_{x\to \infty} 1/x = 0$, the product rule for limits implies $\lim_{x\to \infty} x^2\sin(1/x)\cdot 1/x$ equals $L \cdot 0 = 0$. That is, $\lim_{x\to \infty} x\sin(1/x) = 0$. Then $\lim_{u\to 0+} \sin(u)/u = 0$, a contradiction.

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Although I don't know what you mean by using $u$-substitution for these limits, you are correct to say that these limits do not exist. We can say a little bit more from the point of view of an early calculus student, and a lot more from the perspective of a later calculus student.

Let us take it for granted that $$ \lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{x}{\sin x} = 1,$$ which is similar to knowing that the derivative of $\sin x$ is $\cos x$ and evaluating this at $1$. What this means is that $\sin(x) \approx x$ for really small $x$. It is in this sense that we call the derivative the best linear approximator.

This also means that $\sin \left( \frac{1}{x} \right) \approx \frac{1}{x}$ for really large $x$. And so $x^2 \sin \left( \frac{1}{x} \right) \approx x$, which goes to $\infty$. This works similarly for the other limits. They're all infinite. This can easily be made more rigorous with proper use of the definitions of limits.

From the perspective of a student who knows Taylor's Theorem, this is very quickly done. In particular, we don't need to restrict ourselves to best linear approximators. We can use best cubic approximators, for instance. Then you would know that $$ \sin(x) = x - \frac{x^3}{3!} + \ldots = x + O\left(x^3\right),$$ so that $$ \sin\left( \frac{1}{x}\right) = \frac{1}{x} - \frac{1}{x^3 3!} + \ldots = \frac{1}{x} + O\left(\frac{1}{x^3}\right).$$ Then multiplying by $\sin \frac{1}{x}$ is seen to be almost exactly like multiplying by $\frac{1}{x}$ for large $x$.

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