When is it possible to pass to the limit in the base and the exponent separately?

Question

$$\eqalign{ & \mathop {\lim }\limits_{n \to \infty } {\left( {{{4{n^2}} \over {(2n + 1)(2n - 1)}}} \right)^{1 - {n^2}}} = \mathop {\lim }\limits_{n \to \infty } {\left( {{1 \over {{{(2n + 1)(2n - 1)} \over {4{n^2}}}}}} \right)^{1 - {n^2}}} = \mathop {\lim }\limits_{n \to \infty } {\left( {{1 \over {{{4{n^2} - 1} \over {4{n^2}}}}}} \right)^{1 - {n^2}}} = \mathop {\lim }\limits_{n \to \infty } {\left( {{1 \over {1 - {1 \over {4{n^2}}}}}} \right)^{1 - {n^2}}} = \mathop {\lim }\limits_{n \to \infty } {1 \over {{{\left( {1 - {1 \over {4{n^2}}}} \right)}^{1 - {n^2}}}}} = \mathop {\lim }\limits_{n \to \infty } {1 \over {{{\left( {{{\left( {1 + {1 \over { - 4{n^2}}}} \right)}^{ - 4{n^2}}}} \right)}^{{{1 - {n^2}} \over { - 4{n^2}}}}}}} = \cr & \mathop {\lim }\limits_{n \to \infty } {1 \over {{e^{{{1 - {n^2}} \over { - 4{n^2}}}}}}} = {1 \over {{e^{{1 \over 4}}}}} = {\left( {{1 \over e}} \right)^{{1 \over 4}}} \cr} $$

Is is right to do this?

EDIT:
Please notice that at some point I "converted" the expression $\mathop {\lim }\limits_{n \to \infty } {\left( {1 + {1 \over { - 4{n^2}}}} \right)^{ - 4{n^2}}}$ to $e$, and then kept evaluating the "rest" of the expression.

Why is it legal?

Answer 1

We have $$ \lim_{n \to \infty } \frac{1}{{{\left( {{{\left( {1 + {1 \over { - 4{n^2}}}} \right)}^{ - 4{n^2}}}} \right)}^{{{1 - {n^2}} \over { - 4{n^2}}}}}} = \lim_{n\to\infty} a_n ^{b_n}, $$ where $\lim_{n\to\infty} a_n = e$ and $b_n = -\frac{1-n^2}{-4n^2}$.

Because $\lim_{n\to\infty} b_n$ exists and the function $f(x,y)=x^y$ is continuous at $(x,y)$ when $x=e$, it is acceptable to say $$ \lim_{n\to\infty} a_n ^{b_n} = \left(\lim_{n\to\infty} a_n\right)^{\lim_{n\to\infty} b_n}. $$

Answer 2

A possibly simpler way is to write

$\begin{array}\\ {1 \over {{{\left( {1 - {1 \over {4{n^2}}}} \right)}^{1 - {n^2}}}}} &= {1 \over {{{\left( {1 - {1 \over {4{n^2}}}} \right)}^{1}}}} {1 \over {{{\left( {1 - {1 \over {4{n^2}}}} \right)}^{- {n^2}}}}}\\ &= {1 \over {{{\left( {1 - {1 \over {4{n^2}}}} \right)}}}}{{{\left( {1 - {1 \over {4{n^2}}}} \right)}^{ n^2}}}\\ \end{array} $

and the first term goes to $1$ and the second to $e^{-1/4}$.