Indeterminate Form Example
\[lim_{x \rightarrow a} \frac{x^x-a^a}{a^x-x^a}= \frac{0}{0}\]
is of indeterminate form, but the limit does have a value which can be found with the use of l'Hospital's Rule.\[\begin{equation} \begin{aligned} lim_{x \rightarrow a} \frac{x^x-a^a}{a^x-x^a} &= lim_{x \rightarrow a} \frac{\frac{d}{dx}(x^x-a^a)}{\frac{d}{dx}(a^x-x^a)} \\ &= lim_{x \rightarrow a} \frac{\frac{d}{dx}(e^{ln(x^x)}-a^a)}{\frac{d}{dx}(e^{ln(a^x)}-x^a)} \\ &= lim_{x \rightarrow a} \frac{\frac{d}{dx}(e^{xln(x)}-a^a)}{\frac{d}{dx}(e^{xln(a)}-x^a)} \\ &= lim_{x \rightarrow a} \frac{\frac{d}{dx}(e^{xln(x)}-a^a)}{\frac{d}{dx}(e^{xln(a)}-x^a)} \\ &= lim_{x \rightarrow a} \frac{e^{xln(x)}(lnx+1)}{e^{xln(a)}lna-ax^{a-1}} \\ &= lim_{x \rightarrow a} \frac{x^x(lnx+1)}{a^xlna-ax^{a-1}} \\ &= \frac{a^a(lna+1)}{a^alna-a^a}=\frac{lna+1}{lna-1} \end{aligned} \end{equation}\]
