Question

how to find this

Answer 1

this???

Answer 2

here it iss

Answer 3

help!!

Answer 4

You need to draw it more clearly.

Answer 5

prove whether it exist?

try use one sided limit

Answer 6

is it e^x

Answer 7

is it okay?

Answer 8

e^x or e^(-x)

Answer 9

lim x->0 {(e^-x -e^-1)/x-1}

Answer 10

quick guys PLEASE PLEASE....EVALUATE

Answer 11

wolfram says it is -1/e I'm not sure how it got this though :S sorry

Answer 12

@ang i need to learn steps not answers, that i can check frm back pages
anyway thanks

Answer 13

\[\large\frac{\frac{1}{e^x} - \frac1e}{x-1} =\frac{1}{e}\cdot\frac{e^{-x}\cdot e - 1}{x-1}=-\frac1e \cdot\frac{e^{1-x}-1}{1-x} = -\frac1e\]

Answer 14

these are appearing in comands how to see it in equations...

Answer 15

don't u have 1-1=0 in the denom Ishaan

Answer 16

I am leaving the constant out or -1/e.

http://en.wikipedia.org/wiki/Taylor_series

\[\large \frac{e^{1-x}-1}{1-x} = \frac{1 + \frac{1-x}{1!} + \frac{(1-x)^2}{2!} + \ldots+\frac{(1-x)^{n}}{n!}-1}{1-x} \]

Answer 17

all of these are appearing in "[\large \frac{e^{1-x}-1}{1-x} = \frac{1 + \frac{1-" how to see this in a form of equation ,, Im not able to get it!!!

Answer 18

Refresh!

Answer 19

MY GODDDD
:-D :- D:-D :-D

Answer 20

Wait x->1, your question is wrong if x-> 0 then it's \[\frac{1 - \frac1e}{-1}\]

Answer 21

or \[\frac1e -1\]

Answer 22

its correct perfect

Answer 23

I didn't mean your question is wrong but all I am saying is I was under the impression that x->1 all this time for x->0 you only need to put the limit in, no need of such manipulation

Answer 24

thanks dost , 1 more quesion, please

Answer 25

Done, post it!

Answer 26

x->0 :O

Answer 27

waah U R THE BEST

Answer 28

Do you know differentiation?

Answer 29

yes , L hospital's rule use karun kya

Answer 30

Apply it, twice.

Answer 31

Yes, L'Hospital use karna hai

Answer 32

I dnt know that rule that much, please make me understand that how to decide the no, of required times of difrentiation

Answer 33

kaise pata chalega ki kitni bar karna hai

Answer 34

http://patrickjmt.com/l%E2%80%99hospitals-rule-and-indeterminate-quotients/

Answer 35

x^2 -> 2x -> 2
isliye do baar

Answer 36

I got it YESSS!!!!