What is this limit $\lim_{x \to \infty} (\frac{1}{x} - 1)^x$?

Question

I know that $\lim_{x \to \infty }(1 + \frac{1}{x})^x=e$, but what if it is $\lim_{x \to \infty }( \frac{1}{x}-1)^x$?

Answer 1

The limit does not exist. To see this, first pull out $-1$ from the base of the exponent

$$\left(\frac{1}{x}-1\right)^x=(-1)^x\left(1-\frac{1}{x}\right)^x$$

Now there is a common identity which you may already know of that states $$\lim_{x\to \infty}\left(1+\frac{a}{x}\right)^x=e^a$$

Therefore, the factor of $\left(1-\frac{1}{x}\right)^x$ converges to $e^{-1}$, but $(-1)^x$ does not converge at all, and so the limit you gave does not exist.

To see why the identity holds, its easier to first see that $\lim_{x\to \infty}\left(1+\frac{a}{x}\right)^x=\lim_{y\to 0}\left(1+ay\right)^{1/y}$ by setting $y=1/x$. Then,

$$\lim_{y\to 0}\left(1+ay\right)^{1/y} = L$$

$$\lim_{y\to 0}\frac{1}{y}\log\left(1+ay\right) = \log L$$

Using L'Hopital's rule we see that

$$\lim_{y\to 0}\frac{\log\left(1+ay\right)}{y} = \lim_{y\to 0}\frac{a/\left(1+ay\right)}{1}=a=\log L$$

Thus $L=e^a$.

Answer 2

The limit does not exist if $x \in R$

because $\lim f(x)^{g(x)}$ requirs $f(x)\ge 0$

if we apply this condition on our limit it becomes $$\frac{1}{x}-1 \ge 0$$ $$\frac{1}{x} \ge 1 $$ $$x \le 1 $$

Answer 3

The limit is in the form $(-1)^\infty$ which intuitively does not converge to a value or exist.