What is this limit equal to? $\lim\limits_{D\to\infty}\left(1+\frac1{D^{\,2}}\right)^D$ 
What is this limit equal to? $$\lim\limits_{D\to\infty}\left(1+\frac1{D^{\,2}}\right)^D$$

Does this limit exist? I think that it goes to $1$ because: $$\left(1+\dfrac1\infty\right)^\infty=1^\infty=1$$
Thank you.
 A: $$\lim_{x\to\infty}\left(1+\frac1{x^2}\right)^x=\left(\lim_{x\to\infty}\left(1+\frac1{x^2}\right)^{x^2}\right)^{\lim\limits_{x\to\infty}\frac1x}=e^0$$
as $\displaystyle \lim_{y\to\infty}\left(1+\frac1y\right)^y=e$
A: For any $x\gt0$, we have $1\leqslant\left(1+\frac1{D^2}\right)^D \lt \left(1+\frac xD\right)^D$ 
when $D$ is sufficiently large. Since $\lim\limits_{D\to\infty}\left(1+\frac xD\right)^D=e^x$ and $\lim\limits_{x\to0^+}e^x=1$, by squeezing principle, $\lim\limits_{D\to\infty}\left(1+\frac1{D^{\,2}}\right)^D=1$.
A: No. 
$$\lim_{x\to\infty}\left(1+\frac{1}{x^2}\right)^x=1.$$
HINT :
$$\lim_{x\to\infty}\left(1+\frac 1x\right)^x=e.$$
A: For large values of D, the corresponding Taylor series is  
$1 + 1/D + 1/(2 D^2)$   
More generally, for $n > 1$, the Taylor expansion of $[1+1/D^n]^D$ starts as $1 + 1 / D^{n-1}$, except for the case of $n=1$ as pointed out by Lab.
A: This is an alternative to Lab's method which may be useful in other problems with exponents.
Take logarithms
$$
\ln \left( 1 + \frac{1}{D^2}\right)^D = D \ln \left( 1 + \frac{1}{D^2}\right) \rightarrow D \cdot \frac{1}{D^2} \rightarrow 0$$
So your limit is $e^0=1$.
