Question

Question #92145

Answer 1

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:

`\lim_(n->\infty )(1+(1)/(n))^(n) = e`

So we take yours:

`\lim_(x->\infty )(1-(4)/(x))^(x)`

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

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

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

Facepalm

Answer 2

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

Explanation:

`(1-4/x)^x = e^(ln(1-4/x)^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_(xrarroo)ln(1-4/x)^x = lim_(xrarroo)x ln(1-4/x)` which has form `oo*0`

So we'll rewrite it as

`lim_(xrarroo) ln(1-4/x)/(1/x)` which has form `oo/oo`

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

`lim_(xrarroo) (1/(1-4/x)* 4/x^2)/(-1/x^2) = lim_(xrarroo) (1/(1-4/x)* 4/(-1)) = 1/(1-0) * (-4) = -4`

Therefore, the limit of the original expression is `e^-4`

More generally

For `lim_(xrarra)(f(x))^g(x)` `" "` where `a` may be `+-oo`.

If the initial form of the limit is `1^oo` or `1^-oo`, we note that the logarithm of the expression is

`ln(f(x))^(g(x)) = g(x) ln(f(x))`.

Whose limit has form `oo * 0`. If we rewrite this, we get

`ln(f(x))/(1/(g(x))`, which has limit of form `0/0`.

Now we can try using l'Hospital's rule to find the limit of the logarithm of `(f(x))^g(x)`. 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`.