# Finding the limit of $\frac{\sqrt{x}}{\sqrt{x}+\sin\sqrt{x}}$

How would one find the limit of

$\displaystyle\lim_{x\to 0}\frac{\sqrt{x}}{\sqrt{x}+\sin\sqrt{x}}$

I know I have to use the L'Hospital rule.

$\displaystyle\lim_{x\to 0}\frac{\frac{1}{2}x^{-1/2}}{\frac{1}{2}\frac{1}{\sqrt{x}}+\frac{1}{2}\frac{1}{\sqrt{x}}\cos\sqrt{x}}$

But I find myself stuck

• answer is 1/2 :) Oct 16, 2013 at 16:01
• Set $\sqrt{x}=y$ and compute $\lim_{y\to0^+}\frac{y}{y+\sin y}$, which is way easier. Oct 16, 2013 at 16:02
• You mean $x \to 0^+$, I guess. Or is this a problem about complex numbers? Oct 16, 2013 at 16:47
• @GEdgar no, he just substituted the parameter, one of the most basic tricks for evaluating a limit. And he did mean $y\to0^+$. Oct 16, 2013 at 23:57

$$\lim_{x\to0}\frac{\sqrt{x}}{\sqrt{x}+\sin\sqrt{x}}=\lim_{x\to0}\frac1{1+\frac{\sin\sqrt x}{\sqrt x}}=\lim_{h\to0}\frac1{1+\frac{\sin h}h}$$ Putting $\sqrt x=h\implies x=h^2$

Continuing from where you left off:

Simply cancel the common factor of $\frac {1}{2 \sqrt x}$ from numerator and denominator:

$$\frac{\frac{1}{2\sqrt x}}{\frac{1}{2}\frac{1}{\sqrt{x}}+\frac{1}{2}\frac{1}{\sqrt{x}}\cos\sqrt{x}} = \dfrac 1{1 + \cos \sqrt x}$$ Now evaluate the limit as $x \to 0$. You should arrive at a limit of $\dfrac 12$.

• I see that makes sense to factor out. the 1/2 1/sqrt(x) Oct 16, 2013 at 16:06
• $\ddot\smile$ @SamiBenRomdhane Jun 20, 2014 at 16:23

Hint: taking reciprocal yields $$\frac{\sin(\sqrt{x}) }{\sqrt{x}}+1.$$

\begin{align} \lim_{x\to 0}\dfrac{\sqrt x}{\sqrt x+\sin\sqrt x}&=\lim_{x\to 0}\dfrac{1}{1+\dfrac{\sin\sqrt x}{\sqrt x}},\quad\sqrt x=t\to x=t^2\\ &=\lim_{x\to0}\dfrac{1}{1+\dfrac{\sin t}{t}},\quad\sin t\sim t\\ &=\lim_{x\to0}\dfrac{1}{1+\dfrac{t}{t}}=\dfrac{1}{2} \end{align}

If you know the Taylor series for sine at x=0, I think you don't even have to use L'Hospital's rule.