## Calculus: Early Transcendentals (2nd Edition)

The solution is: $$\lim_{x\to0^+}x^{2x}=1.$$
To solve this limit we will first transform the expression under the limit. We will use the fact that the natural exponential function and the natural logarithmic function are inverse to each other so $a=e^{\ln a}$. This means that $x^{2x}=e^{\ln x^{2x}}=e^{2x\ln x}$. We additionally used the logarithm rule $\ln b^a=a\ln b$. Our limit becomes $$\lim_{x\to0^+}x^{2x}=\lim_{x\to0^+}e^{2x\ln x}.$$ Now we can use the fact that the exponential function is continuous which allows us to exchange its place with the limit: $$\lim_{x\to0^+}x^{2x}=\lim_{x\to0^+}e^{2x\ln x}=e^{\lim_{x\to0^+}2x\ln x}=e^{2\lim_{x\to0^+}x\ln x}=e^{2l},$$ where we have denoted $l=\lim_{x\to0^+}x\ln x$. Now we have to solve this limit. We will do so by using L'Hopital's rule. "LR" will denote "Apply L'Hopital's rule". $$l=\lim_{x\to0^+}x\ln x=\lim_{x\to0^+}\frac{\ln x}{\frac{1}{x}}=\left[\frac{\ln0^+}{\frac{1}{0^+}}\right]=\left[\frac{-\infty}{\infty}\right][\text{LR}]=\lim_{x\to0^+}\frac{(\ln x)'}{\left(\frac{1}{x}\right)'}=\lim_{x\to0^+}\frac{\frac{1}{x}}{-\frac{1}{x^2}}=\lim_{x\to0^+}\frac{-x^2}{x}=\lim_{x\to0^+}-x=-0=0.$$ Putting this back into the original limit we get $$\lim_{x\to0^+}x^{2x}=e^{2l}=e^{2\cdot0}=1.$$