Vtyjkyuk

I know that $lim_x to infty (1 + frac1x)^x=e$, but what if it is $lim_x to infty ( frac1x-1)^x$?

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

$$left(frac1x-1right)^x=(-1)^xleft(1-frac1xright)^x$$

Now there is a common identity which you may already know of that states $$lim_xto inftyleft(1+fracaxright)^x=e^a$$

Therefore, the factor of $left(1-frac1xright)^x$ 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 $lim_xto inftyleft(1+fracaxright)^x=lim_yto 0left(1+ayright)^1/y$ by setting $y=1/x$. Then,

$$lim_yto 0left(1+ayright)^1/y = L$$

$$lim_yto 0frac1ylogleft(1+ayright) = log L$$

Using L'Hopital's rule we see that

$$lim_yto 0fraclogleft(1+ayright)y = lim_yto 0fraca/left(1+ayright)1=a=log L$$

Thus $L=e^a$.

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 $$frac1x-1 ge 0$$
$$frac1x ge 1 $$
$$x le 1 $$

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