Question

Show the following limit:

lim x-> 0

(e^x -1)
--------
x

Answer 1

do you know l'hopitals

Answer 2

I can't use that, but yes

Answer 3

I'm supposed to get e^0

Answer 4

what is e^0?

Answer 5

1, but I don't know how to get it :P

Answer 6

\[e^{0} = 1\]

Answer 7

hm, but you cant use l'hopitals yet though

Answer 8

Using classic algebra and limit laws.

Answer 9

This is indeterminate. There might be a way to rewrite using logarithmic properties but I've forgotten. If I were solving this I would take the derivative of both the numerator and denominator and plug in.

Answer 10

I have a weird idea maybe if we tried rationalizing the numerator @legomyego180 not sure if this would work

\[\frac{ e^{x}-1 }{ x }*(\frac{ e^{x}+1 }{ e^{x}+1 }) = \frac{ e^{2x}-1 }{ x*e^{x}-1 }\]

Answer 11

I went down that route, but it got too complicated. I might have done something wrong.

Answer 12

Actually, I dont think this is indeterminate. I think this DNE.
@Photon336 thats certainly not wrong

Answer 13

@Photon336 where do I go from here?

Answer 14

maybe insert x = 0 evaluate it and see what you get

Answer 15

ok.

Answer 16

is this \[e^x-1 \]

or
e ^{x-1}

Answer 17

I thought that it would just be 0/0, but I didn't look at it hard enough.

Answer 18

I dont think so.

Answer 19

I've never seen a conjugate used with e before but it works

Answer 20

@legomyego180 have you heard of wolfram alpha? You can check the result on there too

Answer 21

yup, I have pro if you want me to post the steps

Answer 22

yeah that would be great

Answer 23

Used L'hopitals but got the same answer

Answer 24

e^(2*0) -1 =

e^0 -1 =

1- 1 =

0

Answer 25

yeah, wow that's interesting

Answer 26

How did I get -1 before? @Photon336

Answer 27

I ended up getting 1

Answer 28

The final answer is 1, but I think I'm going crazy with simple exponent rules >_> for thinking that was -1.

How do I prove the answer is 1 if your method doesn't work? (Unless it does, and I'm going crazy)

Answer 29

*thinking numb. is -1

Answer 30

see what I did above

Answer 31

I still don't understand, I'm getting 0/-1 = 0

Answer 32

@Photon336

Answer 33

check this out. I found this alternative explanation online.




e^x = 1 + x + x^2/2! + x^3/3! + ....

=> e^x - 1= x + x^2/2! + x^3/3! + ....

=> (e^x - 1)/x = 1 + x/2! + x^2/3! + ...

=> lim (x → 0) (e^x - 1) / x

= lim (x → 0) [1 + x/2! + x^2/3! + ...]

= 1.

Answer 34

@Berrymox so my guess here is that as x->0 and get's really really small, all those terms I guess converge to 1

Answer 35

I feel bad for you typing all that out, but I can't use that method either.

Answer 36

*At least, not in my homework

Answer 37

@Berrymox what methods did you learn in class?

Answer 38

just the methods you were doing before with rationalizing and stuff. Basic algebra.

Answer 39

well, we gotta figure out something

Answer 40

I'll just leave it as a graph and take the point off then

Answer 41

use the definition of the derivative assuming you know the derivative of \(e^x\)

\[\lim_{x\to0}\frac{e^x-1}{x}=\lim_{x\to0}\frac{e^x-e^0}{x-0}=\left.\frac{d}{dx}e^{x}\right|_{x=0}=e^0=1\]

Answer 42

fascinating

Answer 43

\[\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=f'(a)\]

here \(f(x)=e^x\) and \(a=0\)