How to calculate a limit without using L'Hôpital's Rule

Question

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 $x=0$.

Answer 1

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 $+\infty$.

Answer 2

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 $\lim_{u \rightarrow 0} \frac{\sin u^3}{u^2}$.