Another Evaluating Limits By Taking Logs Example
\[lim_{x \rightarrow 0} ( \frac{a^x+b^x}{2})^{\frac{1}{x}}\]
take logs.\[ln((\frac{a^x+b^x}{2})^{\frac{1}{x}})= \frac{1}{x} ln(\frac{a^x+b^x}{2})=\frac{ln(a^x+b^x)/2}{x}\]
. This is of Indeterminate Form as \[x \rightarrow \infty\]
(it equals \[\frac{0}{0}\]
) so use l'Hospital's Rule.\[\begin{equation} \begin{aligned} lim_{x \rightarrow 0} \frac{ln(a^x+b^x)/2}{x} &= lim_{ x \rightarrow 0} \frac{\frac{d}{dx} (ln(a^x+b^x)/2)}{\frac{d}{dx}(x)} \\ &= lim_{ x \rightarrow 0} \frac{(a^xlna+b^xlnb)/2}{1} \\ &= (lna+lnb)/2 \\ &=ln(ab)/2 \\ &= ln(ab)^{1/2} \end{aligned} \end{equation}\]
.Then
\[lim_{x \rightarrow 0} ( \frac{a^x+b^x}{2})^{\frac{1}{x}}=e^{ln(ab)^{1/2}}= \sqrt{ab}\]
. 
