$\lim_{x \rightarrow 0} \frac{ \sin{x}}{\sqrt{x}}=?$

I am having trouble evaluating these limits

$$\lim_{x \rightarrow 0} \frac{ \sin{x}}{\sqrt{x}}=?$$

and

$$\lim_{x \rightarrow 0} \frac{ \sin{x}}{x^2}=?$$

I understand L'Hospitals rule and the fact that

$$\lim_{x \rightarrow 0} \frac{ \sin{x}}{x}=1$$

so the first one gave me 0 but the answer is supposedly undefined.

The second one I was told that it was 0.5 but I think it is undefined.

Can someone help?

• Hint. Write $\frac{\sin x}{\sqrt{x}}=\sqrt{x}\frac{\sin x}{x}$ – Dave Sep 12 '14 at 0:38
• Ah, I see. So the limit on the left side of 0 does not exist. That makes sense. So, if the question were $\lim_{x \rightarrow 0+}$ then would the answer be 0? – hyg17 Sep 12 '14 at 0:41
• That depends on the conventions adopted. When a function is undefined to the left of $0$, but has a limit on the right, you'd have to take a hairsplitting look at the definitions an author uses to say whether the limit "exists" or not at $0$. – Dave Sep 12 '14 at 0:45
• Usually when we take limits, we restrict the values of $x$ to the natural domain of the function. So $\lim_{x \rightarrow 0} \sin(x)/\sqrt{x}$ would only consider positive values of $x$. Assuming this, the limit would be zero for the reason given by @Dave. For $\sin(x)/x^2$, the natural domain is the set of all real numbers except $0$, so the limit has the usual meaning without restricting $x$. Note that $$\frac{\sin(x)}{x^2} = \frac{1}{x} \frac{\sin(x)}{x}$$What can you conclude? Hint: the answer is not $1/2$. – Bungo Sep 12 '14 at 0:59
• Sorry hyg17, I've just realized that my comment didn't answer your question properly. The answer is that the limit for $x \to 0^{+}$ would definitely be $0$. My comment addressed whether the limit existed for $x \to 0$. – Dave Sep 12 '14 at 1:03