L'Hopitals Rule?

Question

lim 13 x e^(1/x) −13 x
x -> infinity

Evaluate the limit

Answer 1

First, factor:

13x e^(1/x) −13 x = 13x(e^(1/x) −1)

You need a quotient to apply L'Hopital's Rule, so write this as

[e^(1/x) −1]/[1/(13x)]

Answer 2

lim 13 x e^(1/x) −13 x
x -> infinity
lim 13x(e^1/x -1)
x--> infinity
lim (e^1/x-1)/(1/(13x))
x --> infinity
Above is now in form 0/0 so we can now use L'Hospital's rule

Take derivative of numerator and denominator separately and see if a limit becomes evident. If not continue the process until you can arrive at a definite conclusion.

You should find the limit to be 13.

Answer 3

infinity-infinity form
lim x-> infinity 13 [ e^(1/x) - x/x^2 e^(1/x)] -13
=lim x->infinity 13 [ e^(1/x) - e^(1/x)/x] -13
if you can assume that e^(1/x) /x = 0 as x-> inf (because e^(1/x)-> 1 as x-> inf and 1/x -> 0 as x -> inf)
13[1]-13=0

Answer 4

Use http://en.wikipedia.org/w/index.php?title=L%27H%C3%B4pital%27s_rule&oldid=162670556 and perhaps the product variation?
y= 13x(e^[1/x]-1)