# How do you evaluate [ ( x )^(2x) ] as x approaches 0+?

Oct 31, 2016

I don't think this can be done algebraically. I think you'll need l"Hopital's rule.

#### Explanation:

${x}^{2 x} = {e}^{2 x \ln x}$

Since the exponential function is continuous, we can simplify the problem to find ${\lim}_{x \rightarrow {0}^{+}} x \ln x = L$. (The limit we want will then be ${e}^{2 L}$.)

${\lim}_{x \rightarrow {0}^{+}} x \ln x$ has initial form $0 \cdot \left(- \infty\right)$. This form is indeterminate.

We can apply l'Hopital''s Rule if we can make the form $\frac{0}{0}$ or $\pm \frac{\infty}{\infty}$.

${\lim}_{x \rightarrow {0}^{+}} x \ln x = {\lim}_{x \rightarrow {0}^{+}} \ln \frac{x}{\frac{1}{x}}$ has form $- \frac{\infty}{\infty}$

Applying the rule gets us

$= {\lim}_{x \rightarrow {0}^{+}} \frac{\frac{1}{x}}{- \frac{1}{x} ^ 2} = {\lim}_{x \rightarrow {0}^{+}} \left(- x\right) = 0$.

Therefore, ${\lim}_{x \rightarrow {0}^{+}} \left(2 x \ln x\right) = 0$

We conclude that

${\lim}_{x \rightarrow {0}^{+}} {x}^{2 x} = {\lim}_{x \rightarrow {0}^{+}} {e}^{2 x \ln x}$

$= {e}^{\left({\lim}_{x \rightarrow {0}^{+}} 2 x \ln x\right)}$

$= {e}^{0} = 1$