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I'm unable to solve the following limit: $$\lim_{x \to 0+}\frac{\sin( x \sqrt x )-x \sqrt x}{x\sqrt{x+1}}$$

My attempt: Using L'Hospital rule: $$\frac{ \cos( x \sqrt x ) \frac{3}{2} \sqrt x - \frac{3}{2} \sqrt x } { \sqrt{x+1} + x \frac{1}{2\sqrt{x+1}} }$$ $$\frac{\sqrt{x+1} (\cos( x \sqrt x ) \frac{3}{2} \sqrt x - \frac{3}{2} \sqrt x)}{2x+1}$$ Since $(\sqrt x)'=\frac{1}{2\sqrt x}$ we can see that we will get $\frac{0}{0}$ when $x \to 0$ if my calculations are right.

Alex Provost
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Squidward
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  • Set $x\sqrt x=y$ and use the second of https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%c3%b4pital-rule-or-series-expansion – lab bhattacharjee Dec 18 '20 at 17:38

8 Answers8

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$$\lim_{x \to 0_+}\frac{\sqrt{x+1} (\cos( x \sqrt x ) \frac{3}{2} \sqrt x - \frac{3}{2} \sqrt x)}{2x+1} = \frac{0}{1} = 0$$

Because the denominator resolves to 1.

Andronicus
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Hint: Expand $\sin(x^{3/2})$ near $0$ and you get:

$$\lim_{x \to 0^+}\frac{\sin( x^{3/2})-x^{3/2}}{x\sqrt{x+1}} = \lim_{x \to 0^+}\frac{x^{3/2} - \frac{1}{6}x^{-9/2} + o(x^{13/2}) - x^{3/2}}{x\sqrt{x+1}} = \dots$$

Gabrielek
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I see a lot of answers using Taylor expansions, but this is overkill.

Since $|\sin u|\le |u|$ you get $|\sin u-u|\le 2|u|\quad$ (use for $u=x\sqrt{x}$).

For the denominator $\sqrt{x+1}\ge 1$ when $x\to 0^+$

Therefore $$\left|\dfrac{\sin(x\sqrt{x})-x\sqrt{x}}{x\sqrt{x+1}}\right|\le\left|\dfrac{2x\sqrt{x}}{x}\right|=2\sqrt{x}\to 0$$

Sometime using just rough inequalities is enough to do the job.

zwim
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Dividing numerator and denominator by $x\sqrt x$ gives$$ \lim_{x\to0+}\frac{\frac{\sin (x\sqrt x)}{x\sqrt x}-1}{\sqrt{\frac{x+1}x}}=0$$ because $\lim_{x\to0}\frac{\sin x}x=1$

saulspatz
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The Maclaurin expansion of sine is $\sin(y) = y - y^3/3! + O(y^5)$, so substituting $y = x\sqrt x$ we have $$\sin(x^{3/2}) =x^{3/2} - x^{9/2}/6 + O(x^{15/2}).$$ Hence your limit reduces to

$$\lim_{x \to 0^+} \frac{-x^{9/2}}{6x\sqrt{x+1}} = \lim_{x \to 0^+} \frac{-x^{7/2}}{6\sqrt{x+1}} = 0. $$

Alex Provost
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Use $\sin z \to z-z^3/6$ if $z|<<1$ $$L=\lim_{x \to 0+}\frac{\sin( x \sqrt x )-x \sqrt x}{x\sqrt{x+1}}= \lim_{x \to 0^+} \frac{x^{3/2}-x^{9/2}/6-x^{3/2}}{x\sqrt{x+1}}.$$ $$\implies L=\lim_{x \to 0^+} \frac{-x^{7/2}/6}{\sqrt{x+1}}=0.$$

Z Ahmed
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$$L(x)=\frac{\sin(x^{3/2})-x^{3/2}}{x\sqrt{x+1}}$$ we know that: $$\sin(x)=x-\frac{x^3}{6}+...$$ and so: $$L(x)=\frac{x^{9/2}/6+...}{x\sqrt{x+1}}=\frac{x^{7/2}+...}{6\sqrt{x+1}}$$ now if you take $x\to0$ it is clear that $L(x)\to0$

Henry Lee
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Here is how it can be solved without L'hopital or series expansion :

Observe that for any $ x\in\mathbb{R}^{*} $, we have : $$ \frac{x-\sin{x}}{x^{2}}=\int_{0}^{1}{\left(1-y\right)\sin{\left(xy\right)}\,\mathrm{d}y} $$

Thus, if $ x $ is a real different from $ 0 $, we have : \begin{aligned}\left|\frac{x-\sin{x}}{x^{2}}\right|&=\left|\int_{0}^{1}{\left(1-y\right)\sin{\left(xy\right)}\,\mathrm{d}y}\right|\\ &\leq\int_{0}^{1}{\left(1-y\right)\left|\sin{\left(xy\right)}\right|\mathrm{d}y}\\ &\leq\int_{0}^{1}{\left(1-y\right)\left|x\right|y\,\mathrm{d}y}=\frac{\left|x\right|}{6}\end{aligned}

That means : $$ \lim_{x\to 0}{\frac{x-\sin{x}}{x^{2}}}=0 $$

Thus : \begin{aligned}\lim_{x\to 0^{+}}{\frac{\sin{\left(x\sqrt{x}\right)}-x\sqrt{x}}{x\sqrt{x+1}}}&=\lim_{x\to 0^{+}}{\left(-\frac{x^{2}}{\sqrt{x+1}}\times\frac{x\sqrt{x}-\sin{\left(x\sqrt{x}\right)}}{\left(x\sqrt{x}\right)^{2}}\right)}\\ &=-0\times 0\\ &=0\end{aligned}

CHAMSI
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