Limit $\lim_{x\to\infty}\left(1-\frac{a^2}{x^2}\right)^{x^2}$ I found this example in a textbook, and I understand the author's reasoning and I also reached the same answer using  L’Hôpital’s rule. However, I have two issues:
Firstly: For any finite $a$, then as $x \rightarrow \infty$, $1-\frac{a^2}{x^2} \rightarrow1$, and 1 to the power of anything is 1, so shouldn't the answer be 1?
Secondly: If I try some form of estimation on my calculator by replacing $x^2$ with $10^{99}$, I also get 1 for any $a$ I use.
 A: We are using that
$$\left(1-\frac{a^2}{x^2}\right)^{x^2}=e^{x^2 \ln\left(1-\frac{a^2}{x^2}\right)}$$
and $x^2 \ln\left(1-\frac{a^2}{x^2}\right)$ is an indeterminate form $\infty \cdot 0$ and since the Taylor's expansion for $\ln (1+t)$ as $t\to 0$ is
$$\ln (1+t)=t-\frac12t^2+\frac13 t^3-\frac14t^4+\ldots$$
the result follows.
As an intuitive explanation of the result and  for the indeterminate form $1^{\infty}$, we need to consider that, in that case, while the base tends to $1^-$ the effect of the exponent is to make the quantity smaller towards zero. The final result is indeed a value in between that is $\frac1{e^{a^2}}$. 
That's of course not always the case since all depends upon the rate of convergence to $1$ for the base and the rate of divergence for the exponent. We can indeed obtain any result between $[0,\infty)$ depending on that. That's why we define that an indeterminate form.
For example as $x \to \infty$


*

*$\forall a\in \mathbb{R}\quad \left(1+\frac ax\right)^x\to e^a$
and as a consequence


*

*$\left(1+\frac ax\right)^{x^2}\to \infty \quad a>0$

*$\left(1+\frac ax\right)^{x^2}\to 1 \quad a=0$

*$\left(1+\frac ax\right)^{x^2}\to 0 \quad a<0$
A: Notice that: 
$$\lim_{x \rightarrow \infty} \Bigg(1-\frac{a^2}{x^2}\Bigg)^{x^2}=\lim_{x^2\to\infty}\Bigg(1-\frac{a^2}{x^2}\Bigg)^{x^2}$$
By using the limit substitution rule ($x\to \infty$ means $x^2 \to \infty$) 
We know that this is a standard limit of the form:
$$ \lim_{p\rightarrow \infty}\left( 1+ \frac{b}{p}\right)^p = e^b$$
So with this reasoning we get for $p =x^2$ and $b=-a^2$:
$$ \lim_{x^2\to\infty}\Bigg(1-\frac{a^2}{x^2}\Bigg)^{x^2}=e^{-a^2}$$
Also see: https://www.wolframalpha.com/input/?i=lim_%7Bx%5Cto%5Cinfty%7D%5CBigg(1-%5Cfrac%7Ba%5E2%7D%7Bx%5E2%7D%5CBigg)%5E%7Bx%5E2%7D

As for your method of checking, I would generate a table (maybe in matlab or Mathematica) try for $a=2$, so we should get $e^{-4}\approx 0.0183$.
$$(1-4)^1=-3$$
$$\vdots$$
$$(1-\frac{4}{36})^{36} \approx 0.01440$$
$$(1-\frac{4}{49})^{49} \approx 0.01541$$
For larger values the computer will start rounding and behaving weirdly, Wolfram alpa gives amazing fractions, but only up to some point. It will also be like "ehhh how about no" after some point and start rounding and truncating.
https://www.wolframalpha.com/input/?i=(1-%5Cfrac%7B4%7D%7B49%7D)%5E%7B49%7D

Actually, wolfram alpha is really amazing, it can easily compute:
$$ (1- \frac{4}{1000})^{1000}=0.0181693095355$$
with impeccable accuracy.
https://www.wolframalpha.com/input/?i=(1-%5Cfrac%7B4%7D%7B1000%7D)%5E%7B1000%7D
At some point it WILL approximate:

A: For the sake of simplicity, I will show you a little bit different limit, but It's quite helpful in understanding that your limit is not $1$:
$$x_n:=\left(1+\frac{a}{n}\right)^n$$
We can use Bernoulli's inequaluty to get that
$$x_n \geq 1+n\frac{a}{n}=1+a$$
So the limit is at least $1+a$ (if it exists).
And we don't need $n$ to be a natural number, it can be real as well, but it's much easier to prove the Bernoulli's inequality for natural exponents.
A: I agree with Rebellos on the first point. You can't ignore the exponent and take the limit of the inside first. This reasoning would give you e=1 (since e is a similarly defined limit). On the second point however, I disagree. Plugging in numbers is a completely valid way of estimating a limit. The problem is, calculators are not exact, they'll truncate the actual value to 1, and then raising 1, instead of 1+a, to the "huge number", (where a is some incredibly small number).
