Manipulating the definition of $e$

Question

I know that $\lim\limits_{n\rightarrow \infty}(1+\frac{1}{n})^n=e$

I'm trying to show $\lim\limits_{t\rightarrow \infty}(1+\frac{1}{t^2})^{t^2}=e$

If I write $n=t^2$ then $\lim\limits_{t^2\rightarrow \infty}(1+\frac{1}{t^2})^{t^2}=e$

Can I conclude from this that $\lim\limits_{t\rightarrow \infty}(1+\frac{1}{t^2})^{t^2}=e$?

Could it not imply $\lim\limits_{t\rightarrow\ -\infty}(1+\frac{1}{t^2})^{t^2}=e$ (note that $t$ tends to $-\infty)$

Answer 1

Let's have a look at a $\delta-\epsilon$ proof.

We know that if $\lim_{n\to \infty}(1+1/n)^n =e$, then for any given positive number $\epsilon>0$, there exists a number $N(\epsilon)$ that depends on $\epsilon$ such that

$$\left|\left(1+\frac1 n\right)^n-e\right|<\epsilon$$

whenever $n>N(\epsilon)$.

Now, let's have a look at

$$\left(1+\frac1 {t^2}\right)^{t^2}$$

If we take $t>T(\epsilon)=\sqrt{N(\epsilon)}$, then $t^2>N(\epsilon)$ and this implies that

$$\left|\left(1+\frac1 {t^2}\right)^{t^2}-e\right|<\epsilon$$

So, given $\epsilon>0$, there exists a $T(\epsilon)$ such that

$$\left|\left(1+\frac1 {t^2}\right)^{t^2}-e\right|<\epsilon$$

whenever $t>T(\epsilon)$. By definition, this means that

$$\lim_{t \to \infty} \left(1+\frac1 {t^2}\right)^{t^2}=e$$

Answer 2

Both are true. Note that $$\lim_{\circ\to\infty}\left(1+\frac{1}{\circ}\right)^\circ=e$$ and that $t^2\to\infty$ as $t\to\pm\infty$.