## Calculus: Early Transcendentals (2nd Edition)

Step 1. Transform $$\lim_{x\to a}f(x)^{g(x)}=\lim_{x\to a}e^{g(x)\ln f(x)}=e^{\lim_{x\to a}g(x)\ln f(x)}$$ Step 2. Transform $$g(x)\ln f(x)=\frac{\ln f(x)}{\frac{1}{g(x)}}$$ to get the $\infty/\infty$ form and apply L'Hopital's rule.
The indeterminate limit $\lim_{x\to a}f(x)^{g(x)}$ is always of the form of $1^\infty$. To attack it we can use the following two steps: Step 1. Note that because the natural exponential function and the natural logarithmic function are inverse to each other so we can write $f(x)^{g(x)}=e^{\ln f(x)^{g(x)}}$ which becomes, using the logarithm rule $\ln a^b=b\ln a$, $f(x)^{g(x)}=e^{g(x)\ln f(x)}$. Step 2. Use that the exponential function is continuous so we can write $$\lim_{x\to a}f(x)^{g(x)}=\lim_{x\to a}e^{g(x)\ln f(x)}=e^{\lim_{x\to a}g(x)\ln(x)}.$$ The solution reduces to finding $l=\lim_{x\to a}g(x)\ln f(x)$. This limitis of the form of $\infty\cdot \ln 1=\infty\cdot 0$. To transform it into some expression to which the L'Hopital's rule is directly applicable we use the identity $$g(x)\ln f(x)=\frac{\ln f(x)}{\frac{1}{g(x)}}$$ and this expression when $x\to a$ is of the form of $\infty/\infty$. Now the L'Hopital's rule can be applied for finding $l$ which completes the strategy.