Question

Given F(x)=5-5e^-x, find lim->∞ F(x) using L’Hopital’s Rule.

Answer 1

F(x)=5-5e^-x (x+1)

Answer 2

I'm not sure if that's correct...I also need to show steps but I don't know how to do this problem.

Answer 3

i dint write an answer i only wrote ur qs :)

Answer 4

Oh well then...

Answer 5

\[F(x)=5-5e ^{-x} (x+1) = 5-\frac{ 5(x+1) }{ e^x }\]\[\lim_{x \rightarrow \infty} F(x) =\lim_{x \rightarrow \infty} 5-\frac{ 5(x+1) }{ e^x } = \lim_{x \rightarrow \infty} 5 - \lim_{x \rightarrow \infty} \frac{ 5(x+1) }{ e^x }\]because\[\lim_{x \rightarrow \infty} (F(x) \pm G(x)) = \lim_{x \rightarrow \infty} F(x) \pm \lim_{x \rightarrow \infty} G(x)\]
Using L’Hopital’s Rule we get\[\lim_{x \rightarrow \infty} 5 - \lim_{x \rightarrow \infty} \frac{ 5}{ e^x }\]
Finally evaluate limits
\[5 - 0 = 5 \]

Answer 6

@Lyrae Thank you!!

Answer 7

Yw :)