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