Question

Given F(x) = 5 - (5e^(-x) ) *(x+1), find the limit as x approaches infinity using L'Hopital's Rule.
Can someone give me a step-by-step on this? I'm familiar with L'Hopital's rule, but even when converting this to a fractional problem and following procedure I can't for the life of me figure out out how to do it.

Answer 1

or
\[(5-5e^{-x})(x+1)\]

Answer 2

The first one. Thank you. Being a long-time lurker and short-time poster means I haven't quite gotten used to the equation representation yet

Answer 3

\[5-5e^{-x}(x+1)\] is that the one?

Answer 4

Yup

Answer 5

k, lets rewrite it as
\[5-\frac{5(x+1)}{e^x}\]

Answer 6

now you don't really need l'hopital as \(e^x\) to to infinity much much much faster than \(x\) and therefore the limit of the second part is zero, but you can use it if you like

Answer 7

"Second part"... *facepalm* I understand what I got wrong now... Every way I tackled it got rid of our standalone constant, rather than preserving it and using the Constant Rule for the final limit. i.e. changing 5 so that it can be included in the derivative and effectively destroying it in translation.
Thank you!

Answer 8

yw