Finding $\lim_{x\to 0} \large \frac {\sqrt{x}}{\sin x}$ Using L'Hospitals rule I keep on getting $\frac{0}{0}$... But Im not sure if this is correct?
$$\lim\limits_{x\to 0} \frac{\sqrt{x}}{\sin x}$$
$$\frac{\frac{1}{2}x^{-\frac{1}{2}}}{\cos x}$$
  $$\frac{-\frac{1}{4}x^{-\frac{3}{2}}}{-\sin x}$$
Thanks in advance for any help.
 A: After applying L'Hospital's once, reevaluate: it is no longer in an indeterminate form.
$$\lim_{x\to 0} \frac {\sqrt x}{\sin x} \quad \overset{L'H}= \quad \lim_{x\to 0} \frac 12 \dfrac{1}{\sqrt x\cos x}$$
The right-sided limit (approaching $0$ from the right) exists, and $\lim_{x\to 0^+} \frac 12 \dfrac{1}{\sqrt x\cos x}\to \infty$. But the left-sided limit (approaching $x$ from the left) is not defined, i.e., the limit does not exist.
ADDED: Recall that we can apply L'Hospital if and only if the limit evaluates to an indeterminate form. And after applying it once, with the posted limit, we obtain a form of $1/0$, which is not indeterminate.
A: Hint: $$\lim _{ x\to 0 } \frac { \sqrt { x }  }{ \sin  x } =\lim _{ x\to 0 } \frac { \sqrt { x }  }{ x } \frac { x }{ \sin  x } =\lim _{ x\to 0 } \frac { 1 }{ \sqrt { x }  } $$
A: The given limit is equivalent to:
$$\lim_{x \to 0}  \frac{1}{\sqrt{x}}$$
since:
 $$\lim_{x \to 0} \frac{x}{\sin x} =1$$
we see the limit exist for $x≥ 0$ and
 $$\lim_{x \to 0^+} \frac{1}{\sqrt{x}} \to \infty $$
A: Looking at the problem, you can already say that the limit does not exist. The limit is two-sided, and approaching zero from the left side cannot be done, domain issue of the square root.
