Question

Other method?!? Someone asked me, I said there is not! What do you think?

\[\lim_{x \rightarrow 0} \frac{e-(1+x)^{\frac{1}{x}}}{x}\]

No Hopital, no series expansion

Answer 1

not sure how you would expand in any case

Answer 2

@satellite73

Answer 3

well convert to x goes to infinity as of 1/x

Answer 4

@ikram002p and then?

Answer 5

\(\large \lim_{x \rightarrow \infty } \frac{e-(1+\frac{1}{x} )^{x}}{\frac{1}{x}}\)
haha seems also LH :P

Answer 6

LH can't it is infinity over zero

Answer 7

11 hours ago ??!!! I posted it right now :))

Answer 8

@Catch.me it's 0/0

Answer 9

how did you get e-(1+x)^1/x = 0 ??

Answer 10

mmm so
\(\large \lim_{x \rightarrow \infty } x (e-(1+\frac{1}{x} )^{x} )\) ?

Answer 11

@Catch.me

It is well known\[\lim_{x \rightarrow 0} (1+x)^{\frac{1}{x}}=\lim_{x \rightarrow \infty} (1+\frac{1}{x})^{x}=e\]

Answer 12

hehe so nw we have (infinity . 0)
mmm can we think of squeeze theorem?

Answer 13

That may work also, I don't know :-) Give it a try

Answer 14

im squeezing my head o.o

Answer 15

I got it thanks @mukushla

Answer 16

yes here:
http://www.physicsforums.com/showthread.php?t=250172

Answer 17

and so many other proofs, just google it ;)

Answer 18

i did that

Answer 19

OK then it is 0/0 why don't you use the quotient rule.

Answer 20

quotient rule derivative? well it's Hopital, I look for another one

Answer 21

I'll go back to this later :-) thank you all

Answer 22

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