General rule for limit of $a_n^{b_n}$?

Question

Let $a_n$ and $b_n$ denote two series with well defined limits $a, b \in \mathbb{R}$ for $n\longrightarrow \infty$.

Is it possible to say following?: $$\lim (a_n^{b_n}) = (\lim a_n)^{\lim b_n} = a^b$$

If not, can you give a counterexample?

Edit: Assume that $a$ positive.

Answer 1

That in general does not provide the result.

Assume that $a = -2$ and $b=1/2$ then

$$\lim (a_n^{b_n}) = (\lim a_n)^{\lim b_n} = a^b =(-2)^{1/2} =\sqrt{-2}$$

Does not make any sense in $\Bbb R.$

But If $a>0$ then , there a certain rang from which $a_n >0$ and you can therefore write

$$\lim (a_n^{b_n}) = \lim \exp(b_n\ln a_n =\lim \exp(b\ln a )= a^b $$

If $a_n>0$ and $a=0$ and $b<0$ then $ \ln a_n\to -\infty$ and $$\lim (a_n^{b_n}) = \lim \exp(b_n\ln a_n) =\infty $$

If $a_n>0$ and $a=0$ and $b>0$ then $ \ln a_n\to -\infty$ and $$\lim (a_n^{b_n}) = \lim \exp(b_n\ln a_n) =0 $$

If $a_n>0$ and $a=0$ and $b=0$ we o not know appriory.

Consider $$\left(\frac{1}{n}\right)^\frac{1}{n} \to 1$$ but $$\left(\frac{1}{e^n}\right)^\frac{1}{n} \to e$$

Answer 2

Let $f(x)=e^x$ and $g(x)=\ln{x}$, $\lim\limits_{n\rightarrow+\infty}a_n=A$ and $\lim\limits_{n\rightarrow+\infty}b_n=B$ and we can write $$a_n^{b_n}=e^{b_n\ln{a_n}},$$ where $a_n>0$ and $A>0$ then since $f$ and $g$ are continuous functions, the limit is equal to $$e^{B\ln{A}}=A^B.$$

There are cases like $\lim\limits_{n\rightarrow+\infty}\left(1+\frac{1}{n}\right)^n,$ about which we can think separately.