Can Algebraic Limit Theorem apply to the function in exponential position?

Question

If I try to calculate this limit, $$\lim_{x \to \infty} {\left( 1+\frac{1}{x} \right) ^ { x+1 }}$$ I go through these steps: $$\lim_{x \to \infty} {\left( 1+\frac{1}{x} \right) ^ {{x} \cdot { \frac{x+1}{x} }}}$$

$$e^{\lim_{x \to \infty} {\frac{x+1}{x}}}$$

And I have a question about How the limit can be distributed. So, more precisely, what conditions are needed to make this possible? $$\lim {f(x)}^{g(x)} = \left(\lim f(x) \right)^{\lim g(x)}$$ Is the only condition that $\lim f(x)$ and $\lim g(x)$ converge needed to make this possible, or are there any more conditions?

Answer 1

\begin{align*} \lim\limits_{x \to \infty} f(x)^{g(x)} &= \lim\limits_{x \to \infty} \exp\left(\log(f(x)^{g(x)})\right)\\ &= \lim\limits_{x \to \infty} \exp\left(g(x)\log(f(x))\right)\\ &= \exp\left(\lim\limits_{x \to \infty} g(x)\log(f(x))\right)\\ &= \exp\left(\lim\limits_{x \to \infty} g(x) \cdot \lim\limits_{x \to \infty} \log(f(x))\right)\\ &= \exp\left(\lim\limits_{x \to \infty} g(x) \cdot \log(\lim\limits_{x \to \infty} f(x))\right)\\ &= \ldots \\ &= \lim\limits_{x \to \infty} f(x)^{\lim\limits_{x \to \infty} g(x)}, \end{align*} so what you need is the existence of the limits and $\lim\limits_{x \to \infty} f(x) > 0$.

Answer 2

If $\lim\limits_{x\rightarrow \infty}f(x)=A>0$ and $\lim\limits_{x\rightarrow\infty}g(x)=B$ so $$\lim_{x\rightarrow\infty}f(x)^{g(x)}=A^B.$$ Indeed, since there is $\epsilon>0$ for which $f(x)>0$ for all $|x|>\epsilon,$ $e^x$ and $\ln$ they are continuous functions, we obtain: $$\lim_{x\rightarrow\infty}f(x)^{g(x)}=\lim_{x\rightarrow\infty}e^{g(x)\ln{f(x)}}=e^{\lim\limits_{x\rightarrow\infty}g(x)\ln{f(x)}}=$$ $$=e^{\lim\limits_{x\rightarrow\infty}g(x)\ln\lim\limits_{x\rightarrow\infty}f(x)}= e^{B\ln{A}}=A^B.$$

Answer 3

$exp(x) = e^{x}$ is a continuous function and if we have $a^{b}$ with $a>0$ we can rewritte $a^{b} = e^{b\ln(a)}$ , then if $f(x)>0$ and there exist $\lim_{x \to \infty}g(x)\ln(f(x)) = L$ then there exist $\lim_{x \to \infty}(f(x))^{g(x)} = \lim_{x \to \infty} e^{g(x)\ln(f(x))} = e^{\lim_{x \to \infty} g(x)\ln(f(x))} = e^{L}$

Answer 4

$$ \lim_{x\to\infty} \left( 1+\frac{1}{x} \right) ^{x+1} = $$ $$ \lim_{x\to\infty} \left(\left( 1+\frac{1}{x} \right) ^{x} \cdot \left( 1+\frac{1}{x} \right)\right) = $$ $$ \lim_{x\to\infty} \left( 1+\frac{1}{x} \right) ^{x} \cdot \lim_{x\to\infty}\left( 1+\frac{1}{x} \right) = $$ $$ e \cdot 1 = $$ $$ e $$