$a = \lim_{n \rightarrow \infty} ( 1+ \frac{1}{n^2})^n$ and $b =\lim_{n \rightarrow \infty}( 1+\frac{1}{n})^{n^2}$ 
If $$a = \lim_{n \rightarrow \infty} \Big( 1+ \frac{1}{n^2}\Big)^n\;\;\;\; \text{and}\;\;\;\; b =\lim_{n \rightarrow \infty}\Big( 1+\frac{1}{n}\Big)^{n^2}$$
then  choose the correct option

*

*$a= 1, b=\infty$


*$a=0 ,b=1 $


*$a=\infty, b=1$


*$a =1,b=0$

My works : First option  will correct
$a = \lim_{n \rightarrow \infty} ( 1+ \frac{1}{n^2})^n= \lim_{n \rightarrow \infty}e^\frac{1}{n}=1$
$b =\lim_{n \rightarrow \infty}( 1+\frac{1}{n})^{n^2}=\lim_{n \rightarrow \infty}e^n= \infty$
Is my answer correct or not ?

Generalize:
$$
a_{k,l} = \lim (1+\frac{1}{n^k})^{n^l}\\
$$
So $a=a_{2,1}$ and $b=a_{1,2}$
$$
\ln a_{k,l} = \lim n^l \ln (1+\frac{1}{n^k})\\
= \lim \frac{\ln (1+\frac{1}{n^k})}{n^{-l}}\\
= \lim \frac{\frac{-k/n^{k+1}}{(1+\frac{1}{n^k})}}{-l n^{-l-1}}\\
= \lim \frac{k}{l} \frac{n^{l+1}}{n^{k+1} (1+\frac{1}{n^k})}\\
= \lim \frac{k}{l} \frac{n^{l-k}}{(1+\frac{1}{n^k})}\\
$$
If $l>k$ then $\ln a_{k,l} \to + \infty$. If $l<k$, then $\ln a_{k,l} \to 0$ and therefore $a_{k,l} \to 1$.
