# Question 92145

Apr 5, 2017

See below.

#### Explanation:

You can't [correction tx to @jimh, see other answer] can use L'Hopital here, [but you do not need it].

This is based upon Bernoulli's Interest Compounding equation which is where $e$ comes from originally.

His formula was:

$\setminus {\lim}_{n \to \setminus \infty} {\left(1 + \frac{1}{n}\right)}^{n} = e$

So we take yours:

$\setminus {\lim}_{x \to \setminus \infty} {\left(1 - \frac{4}{x}\right)}^{x}$

and we say, let $x = - 4 y$

$= \setminus {\lim}_{y \to \textcolor{red}{- \setminus \infty}} {\left(1 + \frac{1}{y}\right)}^{- 4 y}$

= \lim_(y->color(red)(-\infty) )((1+(1)/(y))^(y))^(-4) = ???

Facepalm

Apr 5, 2017

If we rewrite it, we can use l'Hospital.

#### Explanation:

${\left(1 - \frac{4}{x}\right)}^{x} = {e}^{\ln {\left(1 - \frac{4}{x}\right)}^{x}}$

Because the exponential function is continuous,

lim_(xrarroo) (1-4/x)^x = e^(lim_(xrarroo)(ln(1-4/x)^x).

So let's focus on the exponent

${\lim}_{x \rightarrow \infty} \ln {\left(1 - \frac{4}{x}\right)}^{x} = {\lim}_{x \rightarrow \infty} x \ln \left(1 - \frac{4}{x}\right)$ which has form $\infty \cdot 0$

So we'll rewrite it as

${\lim}_{x \rightarrow \infty} \ln \frac{1 - \frac{4}{x}}{\frac{1}{x}}$ which has form $\frac{\infty}{\infty}$

Apply l'Hospital, to get the limit of the exponent is

${\lim}_{x \rightarrow \infty} \frac{\frac{1}{1 - \frac{4}{x}} \cdot \frac{4}{x} ^ 2}{- \frac{1}{x} ^ 2} = {\lim}_{x \rightarrow \infty} \left(\frac{1}{1 - \frac{4}{x}} \cdot \frac{4}{- 1}\right) = \frac{1}{1 - 0} \cdot \left(- 4\right) = - 4$

Therefore, the limit of the original expression is ${e}^{-} 4$

More generally

For ${\lim}_{x \rightarrow a} {\left(f \left(x\right)\right)}^{g} \left(x\right)$ $\text{ }$ where $a$ may be $\pm \infty$.

If the initial form of the limit is ${1}^{\infty}$ or ${1}^{-} \infty$, we note that the logarithm of the expression is

$\ln {\left(f \left(x\right)\right)}^{g \left(x\right)} = g \left(x\right) \ln \left(f \left(x\right)\right)$.

Whose limit has form $\infty \cdot 0$. If we rewrite this, we get

ln(f(x))/(1/(g(x))#, which has limit of form $\frac{0}{0}$.

Now we can try using l'Hospital's rule to find the limit of the logarithm of ${\left(f \left(x\right)\right)}^{g} \left(x\right)$. If we find a limit for the log, say $L$ then we can use the exponential function to get the limit of the original expression. In that case the limit will be ${e}^{L}$.