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

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 $\dfrac 12$.
