So as part of a problem, I need to compute the following limit:
$$\lim_{x \to \infty} \left( 1-\frac1{x^2}\right)^{x^2-1} = \frac1e$$
As can be seen, the answer is $\frac1e$. But how do I get there? I entered the limit on Wolfram Alpha, and obtained the answer. But the step-by-step solution was not available. How has this been obtained? Any help would be appreciated.
Thanks in advance.
HINT:
$$\lim_{x\to\infty}\left(1-\dfrac1{x^2}\right)^{x^2-1}=\left(\lim_{x\to\infty}\left[1+\left(-\dfrac1{x^2}\right)\right]^{-x^2}\right)^{\lim_{x\to\infty}\frac{1-x^2}{x^2}}$$
Now use $\lim_{u\to\infty}\left(1+\dfrac1u\right)^u=e$
An identity for $\frac{1}{e}$ is:
$\frac{1}{e} = \lim_{x \to \infty}(1-\frac{1}{x})^x$
If you know that I think it makes sense to try this first because it looks so similar:
$\lim_{x \to \infty}(1 - \frac{1}{x^2})^{x^2-1}$
$= \lim_{x \to \infty}((1 - \frac{1}{x^2})^{x^2} \cdot (1 - \frac{1}{x^2})^{-1})$
$= \lim_{x \to \infty}(1 - \frac{1}{x^2})^{x^2} \cdot \lim_{x \to \infty}(1 - \frac{1}{x^2})^{-1}$ (justified because both limits exist)
$= \lim_{x \to \infty}(1 - \frac{1}{x^2})^{x^2} \cdot 1$
$= \lim_{x \to \infty}(1 - \frac{1}{x^2})^{x^2}$
$= \lim_{x \to \infty}(1 - \frac{1}{x})^{x}$ (justified because $x^2$ is increasing and unbounded)
$= \frac{1}{e}$
Take the natural log first, then evaluate the limit from there: $$\ln\left(1-\frac{1}{x^2}\right)^{x^2-1}=(x^2-1)\ln\left(1-\frac{1}{x^2}\right)$$ As $x\to\infty$, $x^2-1\to\infty$ and $\ln\left(1-\frac{1}{x^2}\right)\to\ln(1)=0$, so their product is an indeterminate form. If we can get this into a ratio of infinity to infinity or zero to zero, we can use L'Hopital's rule. Let's try moving the $x^2-1$ into the denominator: $$\lim_{x\to\infty}\frac{\ln\left(1-\frac{1}{x^2}\right)}{1/(x^2-1)}$$ Now as $x\to\infty$ we get the indeterminate form $\frac{0}{0}$, so we can use L'Hospital's. Take the derivative of the top and bottom. $$\lim_{x\to\infty}\left(\frac{\frac{2}{x^3}}{1-\frac{1}{x^2}}\right)/\left(\frac{-2x}{(x^2-1)^2}\right)$$ Rearranging this we get $$\lim_{x\to\infty}-\left(\frac{\frac{2}{x}}{x^2-1}\right)\cdot\left(\frac{(x^2-1)^2}{2x}\right)=\lim_{x\to\infty}-\frac{1}{x^2}(x^2-1)=-1$$ Now we raise $e$ to the power of this (to undo the log from the beginning), and we get $$\lim_{x\to\infty}\left(1-\frac{1}{x^2}\right)^{x^2-1}=e^{-1}=\frac{1}{e}$$
You can write the limit as
$$\lim_{x \to \infty}\left((1-\frac{1}{x^2})^{1-x^2}\right)^{-1}=\left((\lim_{x \to \infty}(1-\frac{1}{x^2})^{-x^2}\cdot(1-\frac{1}{x^2})\right)^{-1}=(e\cdot1)^{-1}=\frac{1}{e}$$
Here $\lim_{y\to \infty}(1+\frac{1}{y})^y=e$
This solution does not require L'Hopital. However, it does require the following daunting identity. To learn more about it (trust me, once you've used it once, you'll never want to forget it), see this.
$$\lim\limits_{x \to a}{\phi(x)^{\psi(x)}} = e^{\lim\limits_{x \to a}{[(\phi(x)-1)\psi(x)}]}$$
Using this identity...
$\lim\limits_{x \to \infty}{(1-\frac{1}{x^2})^{x^2-1}}$ $= e^{\lim\limits_{x \to \infty}{[((1-\frac{1}{x^2}) - 1)(x^2-1)]}} = e^\theta$
$\theta$ $= \lim\limits_{x \to \infty}{[((1-\frac{1}{x^2}) - 1)(x^2-1)]}$ $= \lim\limits_{x \to \infty}{-\frac{x^2-1}{x^2}} = -1$
$e^\theta = e^{-1} = \frac{1}{e}$
$$\lim\limits_{x \to \infty}{(1-\frac{1}{x^2})^{x^2-1}} = \frac{1}{e}$$
Considering $$A=\left(1-\dfrac1{x^2}\right)^{x^2-1}$$ Take logarithms $$\log(A)=(x^2-1)\log\left(1-\dfrac1{x^2}\right)$$ Consider that for small $t$ Taylor series is $$\log(1-t)=-t-\frac{t^2}{2}+O\left(t^3\right)$$ Replace $t$ by $\dfrac1{x^2}$ which makes $$\log\left(1-\dfrac1{x^2}\right)=-\frac{1}{x^2}-\frac{1}{2 x^4}+O\left(\frac{1}{x^6}\right)$$ $$\log(A)=(x^2-1)\log\left(1-\dfrac1{x^2}\right)=(x^2-1)\left(-\frac{1}{x^2}-\frac{1}{2 x^4}+O\left(\frac{1}{x^6}\right) \right)$$ $$\log(A)=-1+\frac{1}{2 x^2}+O\left(\frac{1}{x^4}\right)$$ Now, Taylor again $$A=e^{\log(A)}=\frac{1}{e}+\frac{1}{2 e x^2}+O\left(\frac{1}{x^4}\right)$$ which shows not only the limit but also how it is approached.
