Right Hand Limits I was told to evaluate the following limit:
$$\lim_{x \to 0^+} x^{x^2}$$
I know that the limit at 0 is equal to one since $0^0 = 1$, but I don't know what the correct way to do a RHL/LHL is. Do I just sub in a value > 0?
 A: Let $$A=\lim_{x \to 0^+} x^{x^2}$$
Now take logarithm both the sides.
$$\ln A= \lim_{x \to 0^+} x^2 \ln x$$
Now you can use L'Hopitals rule :
$$\ln A = \lim_{x \to 0^+} \frac{\ln x}{\frac{1}{x^2}}$$
$$\ln A= \lim_{x \to 0^+} \frac{\frac{1}{x}}{\frac{-2}{x^3}}=\lim_{x \to 0^+} -\frac{x^2}{2} =0$$
$$\ln A =0 \implies A=1$$
A: Whether or not you define $0^0=1$, this has no relevance for the problem at hand, because you are not computing $0^0$, but a limit.
Note that the power $a^x$ (where $x$ is allowed to take any real value) can only sensibly be defined for $a>0$, so the equality
$$
a^x=\exp(x\log a)
$$
(natural logarithm and standard exponential function) holds.
One could also define $0^x=0$ (for $x>0$), but it would only be marginally useful. And it turns out that the two variable function
$$
f(x,y)=x^y
$$
defined for $x>0$ and any $y$ has no limit for $(x,y)\to(0,0)$.
This explained, you should always treat limits of the form
$$
\lim_{x\to c}f(x)^{g(x)}
$$
(two-sided or one-sided) with the following strategy:


*

*compute $\lim_{x\to c}g(x)\log f(x)$

*if the limit in 1 exists and is finite, say $l$, then $\lim_{x\to c}f(x)^{g(x)}=e^l$

*if the limit in 1 exists and is $\infty$, then $\lim_{x\to c}f(x)^{g(x)}=\infty$

*if the limit in 1 exists and is $-\infty$, then $\lim_{x\to c}f(x)^{g(x)}=0$

*if the limit in 1 does not exist, then neither the original limit exists.
In this case,
$$
\lim_{x\to0^+}x^2\log x=0
$$
which is a basic limit, so indeed
$$
\lim_{x\to0^+}x^{x^2}=e^0=1
$$
