Is $ \lim_{n \to \infty} a_n ^{b_n} = e^{\lim_{n \to \infty}(a_n - 1)b_n}$ always true?

Consider $a_n$ and $b_n$ are two sequences which $\lim _{n \to \infty} a_n = 1$ and $\lim _{n \to \infty} b_n = \infty$ . Can we always use this formula ?

$$ \lim_{n \to \infty} a_n ^{b_n} = e^{\lim_{n \to \infty}(a_n - 1)b_n}$$

Also, when can we use this method for functions ?

A famous case is $a_n = 1+ \frac{1}{n}$ and $b_n = n$ . So $\lim_{n \to \infty}(a_n - 1)b_n = 1$ and $a_n ^{b^n} = e^1 = e$

Yes, this formula can always be used.

Let's take a look at it derivation.It will be clear from the derivation that where it can be used.

$$\text{let}~~L= \lim_{n \to \infty} a_n ^{b_n}$$

$$\ln L= \lim_{n \to \infty} b_n \ln a_n$$ $$\ln L= \lim_{n \to \infty} b_n \ln (1+(a_n-1))$$

Since $a_n \to 1 \implies a_n-1 \to 0$ therefore, we can use the fact that : $$\lim_{x \to 0}\dfrac{\ln(1 + x)}{x}=1$$

We get

$$\ln L= \lim_{n \to \infty} b_n \left(\frac{\ln (1+a_n-1)}{a_n-1}\right)(a_n-1)=\underbrace {\lim_{n \to \infty} \left(\frac{\ln (1+a_n-1)}{a_n-1}\right)}_{=1} \cdot \lim_{n \to \infty} b_n(a_n-1) $$ $$ \implies \ln L= \lim_{n \to \infty} b_n (a_n-1)$$ Hence $$L=e ^{\lim_{n \to \infty} b_n (a_n-1)}$$