Calculating the limit $\lim\limits_{x \to 0^+} \frac{\sqrt{\sin x}-\sin\sqrt{ x}}{x\sqrt{x}}$

Question

Calculate $$\lim\limits_{x \to 0^+} \dfrac{\sqrt{\sin x}-\sin \sqrt{x}}{x\sqrt{x}}$$ without use Taylor serie and L'Hôpital.

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$$\lim\limits_{x \to 0^+} \dfrac{\sqrt{\sin x}-\sin \sqrt{x}}{x\sqrt{x}}\cdot\dfrac{\sqrt{\sin x}+\sin \sqrt{x}}{\sqrt{\sin x}+\sin \sqrt{x}}=\lim\limits_{x \to 0^+} \dfrac{\sin x-\sin^2\sqrt{x}}{x\sqrt{x}(\sqrt{\sin x}+\sin \sqrt{x})}$$

now what ?

Answer 1

$$\mathrm L =\lim\limits_{x \to 0^+} \dfrac{\sqrt{\sin x}-\sin \sqrt{x}}{x\sqrt{x}}= \lim\limits_{x \to 0^+} \dfrac{(\sqrt{\sin x}- \sqrt{x}) -(\sin \sqrt{x} - \sqrt{x})}{x\sqrt{x}} \\= \overbrace{\lim\limits_{x \to 0^+}\dfrac{(\sqrt{\sin x}- \sqrt{x})}{x\sqrt{x}}}^{\large \rm L^\prime} - \overbrace{\lim\limits_{x \to 0^+}\dfrac{(\sin \sqrt{x} - \sqrt{x})}{x\sqrt{x}}}^{\large \rm L^{\prime\prime}}$$

Let $y = \sqrt{x}$

Then $${\rm L^{\prime\prime}} = \lim\limits_{x \to 0^+}\dfrac{(\sin \sqrt{x} - \sqrt{x})}{x\sqrt{x}} = \lim\limits_{x \to 0^+}\dfrac{(\sin y - y)}{y^3} = -\dfrac 16$$

For the other limit,

$${\rm L^{\prime}}=\lim\limits_{x \to 0^+}\dfrac{(\sqrt{\sin x}- \sqrt{x})}{x\sqrt{x}} =\lim\limits_{x \to 0^+}\dfrac{(\sin x- x)}{x^3}\dfrac{x\sqrt{x}}{(\sqrt{\sin x} + \sqrt{x})}= \dfrac {-1}6\lim\limits_{x \to 0^+}\dfrac{x}{\left(\sqrt{\dfrac{\sin x}{x}} + 1\right)} = 0$$

Hence our limit is $\mathrm L = \dfrac {1}{6}$.

I used $$\lim_{x \to 0} \dfrac{\sin x - x}{x^3} = \dfrac{-1}6$$ Are all limits solvable without L'Hôpital Rule or Series Expansion ?

Answer 2

The hint.

Prove that for all $x>0$ we have: $$-\frac{1}{6}<\frac{\sin{x}-x}{x^3}<-\frac{1}{6}+\frac{x^2}{120}$$ and from this we obtain:

$$\lim_{x\rightarrow0^+}\frac{\sin{x}-x}{x^3}=-\frac{1}{6}.$$

Let $f(x)=\sin{x}-x+\frac{1}{6}x^3$, where $x>0$.

Thus, $f'''(x)=-\cos{x}+1\geq0$, which gives

$f''(x)>f''(0)=0$, $f'(x)>f'(0)=0$ and $f(x)>f(0)=0$,

which gives a proof of the left inequality.

By the same way we can prove the right inequality.

Answer 3

note that $$ \begin{split} \lim_{x\to0^{+}} \frac{\sin x-\sin^2\sqrt{x}}{x^2}&\overset{t=\sqrt{x}}{=} \lim_{t\to0^{+}} \frac{\sin(t^2)-\sin^2t}{t^4}\\ &=\lim_{t\to0^{+}} \frac{(t^2-\frac{t^6}{6}+\frac{t^{10}}{120}-\ldots)-(t^2-\frac{t^4}{3}+\frac{2t^{6}}{45}+\frac{t^8}{315}-\ldots}{t^4}\\ &=\lim_{t\to0^{+}} \frac{\frac{t^4}{3}-\frac{19t^6}{90}+\frac{t^8}{315}+\ldots)}{t^4}\\ &=\frac{1}{3} \end{split} $$ Hence $$ \lim_{x\to0^{+}} \frac{\sin x-\sin^2\sqrt{x}}{x\sqrt{x}(\sqrt{\sin x}+\sin\sqrt{x})}= \lim_{x\to0^{+}} \frac{\frac{\sin x-\sin^2\sqrt{x}}{x^2}}{\sqrt{\frac{\sin x}{x}}+\frac{\sin\sqrt{x}}{\sqrt{x}}} =\frac{\frac{1}{3}}{\sqrt{1}+1}=\frac{1}{6} $$