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I want to find the following limit using L Hospital's rule: $$ \lim_{x \to \infty} \sqrt{x} \sin( \frac{1}{x}) $$ I know that this can be solved using squeezed theorem from Cal 1: $$ 0 < \sqrt{x}\sin( \frac{1}{x} ) < \frac{1}{x} $$ since $0 < \sin( \frac{1}{x}) < \frac{1}{x} $. What I have done so far is trying to convert it to fraction form $$ \lim_{x \to \infty} \sqrt{x} \sin( \frac{1}{x}) = \lim_{x \to \infty} \frac{\sin( \frac{1}{x})}{\frac{1}{\sqrt{x}} }$$ But what next?

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  • $\begingroup$ It is much simpler to use equivalents. $\endgroup$ – Bernard Jun 7 at 16:41
  • $\begingroup$ ... then check if the fraction satisfies l'Hopital's hypotesis, calculate and see if it works. $\endgroup$ – Saucy O'Path Jun 7 at 16:44
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Your idea of rewriting it that way is good when you have limit like $$ \lim_{x \to \infty} x^2\sin\bigg( \frac{1}{x} \bigg) $$ because you can then make the substitution $u = \frac{1}{x}$ However, for this problem, you should write it first as $$ \lim_{x \to \infty} \sqrt{x}\sin\bigg( \frac{1}{x} \bigg) = \lim_{u \to \infty} u\sin\bigg( \frac{1}{u^2} \bigg) $$ by making the substitution $u = \sqrt{x}$ . Then now you have $$ \lim_{u \to \infty} u\sin\bigg( \frac{1}{u^2} \bigg) = \lim_{u \to \infty} \frac{ \sin\bigg( \frac{1}{u^2} \bigg)}{\frac{1}{u}} =^{L'H} \lim_{u \to \infty} \frac{\frac{-2\cos(\frac{1}{u^2})}{u^3} }{\frac{-1}{u^2} } = \frac{2\cos\bigg( \lim_{u \to \infty} \frac{1}{u^2}\bigg) }{\lim_{u \to \infty} (u) } = \frac{2}{\infty} = 0 $$

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    $\begingroup$ So is it true that $\lim_{x \to \infty} x^n\sin(\frac{1}{x}) = 0$ for all $n<1$ then? $\endgroup$ – MathStudent Jun 7 at 17:43
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Consider the Taylor series expansion for $\sin(x)$. $$\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}$$ For $x^{-1}$, this series is $$\frac{1}{x} - \frac{1}{3! *x^3} + \frac{1}{5! * x^5} - \ldots$$ This is equivalent to $O\Big(\frac{1}{x}\Big)$ (meaning $\sin(\frac{1}{x}) \to \frac{1}{x}$ as $x \to \infty$). So rewrite your equation as $$\lim_{x \to \infty} \sqrt{x} *O\Big(\frac{1}{x}\Big) = O\Big(\frac{1}{\sqrt{x}}\Big)$$ Which goes to $0$ as x goes to infinity.

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  • $\begingroup$ Good technique as well. $\endgroup$ – user209663 Jun 7 at 16:54
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l'hopital method is applicable only for those limits which are indeterminate . this limit can easily be calculated using elemantry algebra enter image description here

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