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


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

now what ?

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  • $\begingroup$ I added some non-math text to the title and removed the d in \dfrac. The first so that the title may be right-clicked property in the front page, and the second to make it conform to the rest of the front page. $\endgroup$ – Arthur Sep 19 '17 at 9:04
  • $\begingroup$ See math.stackexchange.com/questions/387333/… $\endgroup$ – lab bhattacharjee Sep 19 '17 at 9:18
  • $\begingroup$ Perhaps Euler equation would help. $\endgroup$ – Gabriel Sandoval Oct 20 '17 at 1:52
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$$\mathrm L =\lim\limits_{x \to 0^+} \dfrac{\sqrt{\sin x}-\sin \sqrt{x}}{x\sqrt{x}}= \lim\limits_{x \to 0^+} \dfrac{(\sqrt{\sin x}- \sqrt{x}) -(\sin \sqrt{x} - \sqrt{x})}{x\sqrt{x}} \\= \overbrace{\lim\limits_{x \to 0^+}\dfrac{(\sqrt{\sin x}- \sqrt{x})}{x\sqrt{x}}}^{\large \rm L^\prime} - \overbrace{\lim\limits_{x \to 0^+}\dfrac{(\sin \sqrt{x} - \sqrt{x})}{x\sqrt{x}}}^{\large \rm L^{\prime\prime}}$$

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 ?

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  • $\begingroup$ From where did you get $\lim_{x\to0} (\sin x-x)/x^3 = -1/6$? $\endgroup$ – skyking Sep 19 '17 at 9:14
  • $\begingroup$ @skyking I have linked it. $\endgroup$ – user8277998 Sep 19 '17 at 9:15
  • $\begingroup$ That was elaborate, but that's probably what one gets for not using the proper tools (ie the forbidden ones for this question). $\endgroup$ – skyking Sep 19 '17 at 9:21
  • $\begingroup$ @skyking I like doing this way though. :) $\endgroup$ – user8277998 Sep 19 '17 at 9:25
  • $\begingroup$ The link you gave only says that if the limit of $(\sin x-x) /x^{3}$ exists then it must be $-1/6$. The limit is not possible to evaluate without differentiation /integration. But still the technique used is my favorite. I gave the same technique in my answer but deleted it after I saw yours. +1 $\endgroup$ – Paramanand Singh Sep 19 '17 at 18:03
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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.

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  • $\begingroup$ Except you haven't proved the first inequality which has to be done without using Taylor or l'Hospital. $\endgroup$ – skyking Sep 19 '17 at 9:20
  • $\begingroup$ @skyking See please better my solution. I did not use these things. $\endgroup$ – Michael Rozenberg Sep 19 '17 at 9:22
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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 $$ \lim_{x\to0^{+}} \frac{\sin x-\sin^2\sqrt{x}}{x\sqrt{x}(\sqrt{\sin x}+\sin\sqrt{x})}= \lim_{x\to0^{+}} \frac{\frac{\sin x-\sin^2\sqrt{x}}{x^2}}{\sqrt{\frac{\sin x}{x}}+\frac{\sin\sqrt{x}}{\sqrt{x}}} =\frac{\frac{1}{3}}{\sqrt{1}+1}=\frac{1}{6} $$

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    $\begingroup$ Without tailor series sais the question $\endgroup$ – Marios Gretsas Sep 19 '17 at 9:23

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