I want to know the following is true : If $$ c,\ d\in {\bf R},\ \lim_{x\rightarrow 0} f(x)=c>0,\ \lim_{x\rightarrow 0} g(x) =d>0$$ then $$ \lim_{x\rightarrow 0} f(x)^{g(x)} = c^d$$
In calculus book such formula cannot be found.
Consider the problem : $$\lim_{x\rightarrow 0} (1+\sin\ 4x)^{\cot\ x} $$
To find a limit, we must use ${\rm log}$ and L'Hospital. But some student suggests that $$ \lim_{x\rightarrow 0} (1+\sin\ 4x)^{\frac{1}{\sin\ 4x} \frac{\sin\ 4x}{\sin\ x}\cos\ x}=e^4$$
This argument is clear. But I know that it is informal. Is there a minus point in such way ?
Best Answer
Assuming $f(x)>0$ for some open set surrounding $x_0$, we know that because the natural logarithm is continuous on $(0,\infty)$ that \begin{align} \ln \left( \lim_{x\rightarrow x_0}{f(x)^{g(x)}}\right) & =\lim_{x\rightarrow x_0}{\ln \left(f(x)^{g(x)}\right)} \\ & =\lim_{x\rightarrow x_0}{g(x)\ln f(x)} \\ & =\left(\lim_{x\rightarrow x_0}{g(x)} \right)\left(\lim_{x\rightarrow x_0}{\ln f(x)}\right) \\ & =\left(\lim_{x\rightarrow x_0}{g(x)}\right)\ln \left( \lim_{x\rightarrow x_0}{f(x)}\right) \\ & =\ln \left[\left( \lim_{x\rightarrow x_0}{f(x)}\right)^{\left( \lim_{x\rightarrow x_0}{g(x)}\right)}\right] \end{align}
Then, because the natural logarithm is injective, we find that $$\lim_{x\rightarrow x_0}{f(x)^{g(x)}}=\left( \lim_{x\rightarrow x_0}{f(x)}\right)^{\left( \lim_{x\rightarrow x_0}{g(x)}\right)}$$