8
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ answer is 1/2 :) $\endgroup$
    – Manoj
    Oct 16, 2013 at 16:01
  • 3
    $\begingroup$ Set $\sqrt{x}=y$ and compute $\lim_{y\to0^+}\frac{y}{y+\sin y}$, which is way easier. $\endgroup$
    – egreg
    Oct 16, 2013 at 16:02
  • $\begingroup$ You mean $x \to 0^+$, I guess. Or is this a problem about complex numbers? $\endgroup$
    – GEdgar
    Oct 16, 2013 at 16:47
  • $\begingroup$ @GEdgar no, he just substituted the parameter, one of the most basic tricks for evaluating a limit. And he did mean $y\to0^+$. $\endgroup$ Oct 16, 2013 at 23:57

5 Answers 5

12
$\begingroup$

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

$\endgroup$
10
$\begingroup$

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 $\dfrac 12$.

$\endgroup$
2
  • $\begingroup$ I see that makes sense to factor out. the 1/2 1/sqrt(x) $\endgroup$ Oct 16, 2013 at 16:06
  • $\begingroup$ $\ddot\smile$ @SamiBenRomdhane $\endgroup$
    – amWhy
    Jun 20, 2014 at 16:23
10
$\begingroup$

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

$\endgroup$
0
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.