How to prove this limit properly?

Question

If $\lim_\limits{n \to \infty} a_n=1$ and $\lim_\limits{n \to \infty} b_n=\infty$ then $\lim_\limits{n \to \infty} (a_n)^{b_n}=\exp{(\lim_\limits{n \to \infty} (a_n-1)b_n)}$

I have seen a proof, where $\log{(x+1)}=x+o(x)$ were used. But if I used $\log{(x+1)}=x-\frac{x^2}{2}+o(x)$, i would not get the same answer.

Answer 1

HINT.- $$(a_n)^{b_n}=\left(1+(a_n-1)\right)^{\dfrac{b_n(a_n-1)}{a_n-1}}=\left(1+(a_n-1)\right)^{\frac{1}{a_n-1}\cdot b_n(a_n-1)}$$ Thus $$(a_n)^{b_n}\to e^{\lim_{n\to\infty} b_n(a_n-1)}$$ The existence of the limit depends on $\lim_{n\to\infty} b_n(a_n-1)$.

Answer 2

You seem to be asking for a proof of something that is not true. If $a_n = 1+1/n$ for all $n,$ and $b_n = n$ for odd $n$ and $b_n = n^2$ for even $n,$ then $a_n \to 1,$ and $b_n \to \infty.$ But $\lim a_n^{b_n}$ does not exist: As $n\to \infty$ through odd $n,$ $a_n^{b_n} \to e.$ And if $n\to \infty$ through even $n,$ then $a_n^{b_n} \to \infty.$