Question

find the limit as x approaches infinity,
-(x+1) (e^(1/(x+1))-1)

Answer 1

help please

Answer 2

are you allowed to use L'hops rule ?

Answer 3

yes

Answer 4

my problem is how to differentiate those numbers

Answer 5

\[\large \lim \limits_{x\to \infty }-(x+1) (e^{1/(x+1)}-1) = \lim \limits_{x\to \infty }- \dfrac{ (e^{1/(x+1)}-1)}{1/(x+1)} \]

Answer 6

substitute \(\large t = 1/(x+1) \)

as \(\large x \to \infty\),
\(\large t \to 0\)

right ?

Answer 7

yes

Answer 8

the limit changes to

\[\large \lim \limits_{t\to 0 }- \dfrac{ e^{t}-1}{t} \]

Answer 9

since its of form 0/0
you can apply L'hops

Answer 10

\[\large \lim \limits_{t\to 0 }- \dfrac{ e^{t}-1}{t} ~\stackrel{LH}{=}~ \lim \limits_{t\to 0 }- \dfrac{ e^{t}}{1} \]

Answer 11

are we letting t=1/(x+1)?

Answer 12

yes

Answer 13

okay

Answer 14

so after getting l\[\lim_{t \rightarrow 0}\frac{- e ^{t} }{ 1}\]
do we substitute t=1/(x+1) back?

Answer 15

nope, x's are gone
the final answer will be a numeral value

just pluging t = 0 and evaluate

Answer 16

okay

Answer 17

\[\large \lim_{t \rightarrow 0}\frac{- e ^{t} }{ 1} = -e^0 = -1\]

Answer 18

will the answer be -1?

Answer 19

thats it!

Answer 20

okay

Answer 21

thank you

Answer 22

you're welcome!

Answer 23

so it means anytime you have a question like this, you can represent the exponent by a variable and differentiate.