$\limsup_{n\to\infty}{a_n}^{b_n}= \limsup_{n\to\infty}{a_n}^{\lim\limits_{n\to\infty}b_n}$ under certian conditions.

Question

I have a proof that requires the following justification. So, I decided to recast it in the following form:

If $\lim\limits_{n\to\infty}b_n$ exists and $\limsup_{n\to\infty}{a_n}=a>0$, then $\limsup_{n\to\infty}{a_n}^{b_n}= \limsup_{n\to\infty}{a_n}^{\lim\limits_{n\to\infty}b_n}$

My proof

\begin{align} \limsup_{n\to\infty}{a_n}^{b_n}&=\limsup_{n\to\infty}e^{\ln{a_n}^{b_n}}\\&=\limsup_{n\to\infty}e^{b_n\ln{a_n}}\\&=e^{\limsup_{n\to\infty}(b_n\ln{a_n})}\\&=e^{\lim\limits_{n\to\infty}b_n\limsup_{n\to\infty}\ln{a_n}}\\&=e^{\lim\limits_{n\to\infty}b_n\ln{(\limsup_{n\to\infty}a_n)}}\\&=e^{\lim\limits_{n\to\infty}b_n\ln{a}}\\&=e^{\ln{a}^{\lim\limits_{n\to\infty}b_n}}\\&={a}^{\lim\limits_{n\to\infty}b_n}\\&={\limsup_{n\to\infty}{a_n}}^{\lim\limits_{n\to\infty}b_n}\end{align} Please, is this correct? And is there any other proof?

Answer 1

It's correct because $f(x)=e^x$ and $\ln$ are continuous functions.

For all continuous in $a$ function $f$ if $\lim\limits_{n\rightarrow+\infty}a_n=a$ then: $$\lim_{n\rightarrow+\infty}f(a_n)=f\left(\lim_{n\rightarrow+\infty}a_n\right)=f(a).$$

Answer 2

Your proof seems a bit more complicated than it should be. Since $a^b$ is continuous for all $a,b\in \mathbb{R}$, we have that $\limsup_{n\rightarrow\infty}a_n^{b_n} = \limsup_{n\rightarrow\infty}a_n^{\limsup_{n\rightarrow\infty} b_n} $. However, since $\lim_{n\rightarrow\infty} b_n$ exists, we have $\limsup_{n\rightarrow\infty}b_n=\lim_{n\rightarrow\infty}b_n$, from which your claim follows.