Question

A limits question...

Answer 1

\[\lim_{h \rightarrow 0}\frac{ e^{2 + h} +e^2}{ h }\]

Answer 2

The limit as it is would become 0/0, so I think I need to use L'Hopital's rule on it, but I'm not sure what to do with e^(2+h).

Answer 3

i am quite sure there is a negative sign between the terms in the numerator

Answer 4

Oh yeah, you're right, sorry. It is negative.

Answer 5

\(\Large x^{m+n} = x^m \times x^n\)

Answer 6

then factor out e^2 from numerator

Answer 7

\[\lim_{h \rightarrow 0}\frac{ e^{2+h} - e^2 }{ h }\]

Answer 8

\(\Large e^{2+h} = e^2 \times e^h\)

Answer 9

So now I get \[=\frac{ e^2(e^h - 1) }{ h }\]

Answer 10

e^2 is a constant as e is a constant
so you can throw it out of the limit

Answer 11

for e^h-1/h,
whether you use the standard formula or
L'Hopital's rule, your choice.

Answer 12

Using L'Hopital's then, I get (e^h)/1, which is then 1/1. So the answer is then e^2.

Answer 13

correct! :)

Answer 14

Excellent, I got it. Thanks again. :D

Answer 15

welcome ^_^