Question #92145

2 Answers
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:

#\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

Apr 5, 2017

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#.