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I have to calculate the limits $$\lim_{x\rightarrow 0}\frac{\sin(x\sqrt{x})}{x}\;\;\;\text{and}\;\;\; \lim_{x\rightarrow0}\frac{\sin\sqrt{x}}{x}$$ without using L'Hôpital's Rule, and I can't see how. Calculating those derivatives would help me determine if the two functions $x\mapsto\sin\sqrt{x}$ and $x\mapsto\sin(x\sqrt{x})$ are differentiable at $x=0$.

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    $\begingroup$ You can use Taylor's theorem to get $\sin(x) = x + O(x^3)$ as $x \to 0$. So $$\frac{\sin(x\sqrt{x})}{x} = \sqrt{x} + O(x^{3.5})$$ and $$\frac{\sin(\sqrt{x})}{x} = x^{-1/2} + O(x^{1/2}).$$ $\endgroup$
    – Mason
    Jan 17, 2022 at 2:13
  • $\begingroup$ @Mason Taylor series use derivatives, so kinda are using L'Hopital implicitly. $\endgroup$ Jan 17, 2022 at 6:19
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    $\begingroup$ @Acccumulation Taylor's theorem does not use L'Hopital's rule. The method of proof is different. Admittedly, the proof of Taylor's theorem is a bit more elaborate than the proof of the version of L'Hopital's rule needed here, but I like Taylor's theorem because it gives detailed quantitative information. $\endgroup$
    – Mason
    Jan 17, 2022 at 6:24
  • $\begingroup$ @Mason. I like your comment (I am in love with Taylor series for 65+ years). Moreover, we could say that L'Hopital's rule is a byproduct of Taylor expansion. Cheers ;-) $\endgroup$ Jan 17, 2022 at 8:57

2 Answers 2

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HINT

The result of the first limit is given by

\begin{align*} \lim_{x\to 0^{+}}\frac{\sin(x\sqrt{x})}{x} & = \lim_{x\to 0^{+}}\frac{\sqrt{x}\sin(x\sqrt{x})}{x\sqrt{x}} = 0 \end{align*}

Can you justify why it is so?

On the other hand, we can rewrite the second limit as follows \begin{align*} \lim_{x\to 0^{+}}\frac{\sin(\sqrt{x})}{x} & = \lim_{x\to 0^{+}}\frac{1}{\sqrt{x}}\left(\frac{\sin(\sqrt{x})}{\sqrt{x}}\right) \end{align*}

As $x$ approaches $0$ from the right, the second term tends to one and the first converges to $+\infty$.

Since $1\times +\infty = +\infty$, we conclude it converges to $+\infty$.

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You should look at how $\lim \frac{\sin x}x$ is calculated, and go from there. Also, it's probably easier to use the substitution $u=\sqrt x$. Then you have $\lim_{u \rightarrow 0} \frac{\sin u^3}{u^2}$.

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