Question

lim x approaches 0 [(e^x) - (e^sinx)]/x-sinx

Answer 1

Indeterminate form!

Answer 2

Have you heard of l'Hospital?

Answer 3

nope :(
not yet

Answer 4

In that case this is tough.

Answer 5

u can say that again ;)
thanks though!!

Answer 6

Though it almost looks like a derivative.

Answer 7

IT looks like f'(sin(x)) where f(x) = e^x

Answer 8

which if you think of it that way, then it would be e^sin(x)

Answer 9

Remember that:

f'(a) = [f(x) - f(a)] / (x-a)

Answer 10

as x to a

Answer 11

heyy thats a good way to look at it =D
thanks a lot !!!

Answer 12

I dont think im gettin anywhere with this though :( !!

Answer 13

Okay, so what is a in our case?

Answer 14

What would f(x) be? what would f(a) be?

Answer 15

f(x) would be e^x
f(a) would be e^a
so we have a as Sinx

Answer 16

Okay, so we would say that:

lim x to a [e^x - e^a] / (x-a) = f'(a) = e^a

Answer 17

Likewise
lim x to sin(x) [e^x - e^sin x] / (x-sin x) = e^sin x

Answer 18

yes !!!! =D
hmm but what next ??

Answer 19

If we say that sin x = 0

Answer 20

so would the answer be e^sinx
so that would be e^0 = 1
due to the limit x tends to 0

Answer 21

We could say
lim x to 0 [e^x - e^sin x] / (x-sin x) = e^0

Answer 22

so the answer is 1
right??

Answer 23

Hmm, there is a small problem though...

Answer 24

which is??
:O

Answer 25

if we set sin x to 0, then we also are setting x to be 0 + npi. It is no longer approaching it, technically.
Though I'm not sure this is breaking any limit rules per se. I'm unsure.

Answer 26

@hartnn What do you think?

Answer 27

Is it even reasonable to have a limit where x to f(x)?

Answer 28

ohh that wouldnt be a problem i guess.. coz we gotta find the value so that wouldnt be going overboard :P

Answer 29

i hope my math teacher would buy the idea ;)

Answer 30

for example, if you said limit x to 2x.... you are essentially saying x to 0 since that is the only place where 2x = x

Answer 31

yaa i see your point!

Answer 32

To say x goes to x+1 would just plain not work.

Answer 33

" lim x to a [e^x - e^a] / (x-a) = f'(a) = e^a"
only if 'a' is constant and not a function of x.
though it was a very good try, i don't think its a legitimate one.
thinking on other ways to solve...

Answer 34

ohh!!
thanks I compleetly forgot that :O

Answer 35

Why does a have to be constant again? Because of chain rule?

Answer 36

so you are saying for applying lim x->0 e^sinx we go for e^f(x)

Answer 37

The only thing I can think of is some magical algebra trick of squeeze theorem.

Answer 38

:P heheheee you just red my mind

Answer 39

i have something in mind, but can't draw or use latex...
let me try text...

Answer 40

[{(e^x) -1}- {(e^sinx)-1}]/ (x sin x) times (x sin x)/(x-sin x)

ddoes that make sense ?
donno whether that will work, just a try

Answer 41

trying to use this standard limit

lim x->0 (a^x-1)/x = ln a

Answer 42

well i get the part where we subract 1 from both entities of the numerator
but just to be sure..

Answer 43

[(1/sin x) (e^x-1)/x - (1/x) (e^sin x -1)/ sinx ] / (xsinx) times ....

Answer 44

are we multiplying and dividing the denominator by xsinx??

Answer 45

well i think the denominator is x-sinx!!

Answer 46

yes.....now i fell that might not work though...

Answer 47

yeah, x-sin x is still there,
.....times x sin x/(x- sin x)

Answer 48

ohh ...
if only i could get rid of the denominator becoming zero!!

Answer 49

if only I was right

Answer 50

;)

Answer 51

well, you're getting correct final answer though....

Answer 52

whatt !!!! :O
so the answer is 1 ??

Answer 53

sure it is.

Answer 54

hmm... now i have the option of reverse step building!!!
how..... what would make it 1 .. hmmm.....

Answer 55

[(1/sin x) (e^x-1)/x - (1/x) (e^sin x -1)/ sinx ] / (xsinx) times x sin x/(x- sin x)

=[ lim x->0 (1/sin x) lim x->0 (e^x-1)/x - lim x->0 (1/x) lim x->0 (e^sin x -1)/ sinx times lim x->0 x sin x/(x- sin x)


just distributing the limits

Answer 56

=[ lim x->0 (1/sin x) (1) - lim x->0 (1/x) (1) times lim x->0 x sin x/(x- sin x)

= lim x->0 (1/sin x - 1/x ) times (x sin x)/(x -sin x)

Answer 57

=lim x-> 0 (x-sin x)/x sin x time (x sin x)/(x -sin x)

lim x-> 0 1
= 1

Answer 58

does that look confusing? i wish i could use latex/draw

Answer 59

ohh we are separately applying limit to E to eliminate it first and then evaluating??

Answer 60

yes, thats my idea, not sure if i did any illegal step there, but final answer comes out to be 1...

Answer 61

well we take lim x-> 0 (x-sinx)/xsinx * xsinx/(x-sinx)
to get lim x->0 (x-sinx)/(x-sinx) to get 1 right??

Answer 62

yeah, the numerator and denominator just cancels each other to leave out 1

Answer 63

but exactly how is lim x->0 (e^x-1)/x turn out to be 1??
same for im x->0 (e^sinx -1)/sinx

Answer 64

lim x->0 (a^x-1)/x = ln a

Answer 65

so, lim x-> 0 e^x -1 /x = ln e =1

Answer 66

lim x-> 0 e^ sin x -1 / sin x

put y= sin x
as x -> 0, sin x -> 0 , y-> 0

lim y->- e^y -1/ y = ln e = 1

Answer 67

OHHHHHH yeah now i get it !!!


SWEEETT =D
thanks a lot Hartnn
and you tooo wio =D

Answer 68

most welcome ^_^