# How to use L Hospital's rule for $\lim_{x \to \infty} \sqrt{x} \sin( \frac{1}{x})$

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?

• It is much simpler to use equivalents. – Bernard Jun 7 at 16:41
• ... then check if the fraction satisfies l'Hopital's hypotesis, calculate and see if it works. – Saucy O'Path Jun 7 at 16:44

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 $$\lim_{u \to \infty} u\sin\bigg( \frac{1}{u^2} \bigg) = \lim_{u \to \infty} \frac{ \sin\bigg( \frac{1}{u^2} \bigg)}{\frac{1}{u}} =^{L'H} \lim_{u \to \infty} \frac{\frac{-2\cos(\frac{1}{u^2})}{u^3} }{\frac{-1}{u^2} } = \frac{2\cos\bigg( \lim_{u \to \infty} \frac{1}{u^2}\bigg) }{\lim_{u \to \infty} (u) } = \frac{2}{\infty} = 0$$
• So is it true that $\lim_{x \to \infty} x^n\sin(\frac{1}{x}) = 0$ for all $n<1$ then? – MathStudent Jun 7 at 17:43
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.