Question

Another limit

Answer 1

\[\lim_{x \rightarrow 0-} \left( 1-\cos x\right)^{1/x}\]

Answer 2

\[\lim_{x\to0^-}(1-\cos x)^{1/x}=\exp{\Bigg(\ln\Big(\lim_{x\to0^-}(1-\cos x)^{1/x}}\Big)\Bigg)\]
Continuity of the natural log function gives you
\[\exp{\Bigg(\lim_{x\to0^-}\ln\Big((1-\cos x)^{1/x}}\Big)\Bigg)\]
Property of logarithms gives you
\[\exp{\Bigg(\lim_{x\to0^-}\frac{1}{x}\ln(1-\cos x)}\Bigg)\\
\exp{\Bigg(\lim_{x\to0^-}\frac{\ln(1-\cos x)}{x}}\Bigg)\]
You get \(\dfrac{-\infty}{-\infty}\) by evaluating directly, so L'Hopital;s rule comes into play once again.

Answer 3

ok i did that

Answer 4

that gives u sinx/(1-cosx) right?

Answer 5

\[\lim_{x \rightarrow 0^-} \left( 1-\cos x\right)^{1/x}=\lim_{x \rightarrow 0^+} \left( 1-\cos x\right)^{-1/x}=\lim_{x \rightarrow 0^+} \left(\frac{1}{ 1-\cos x}\right)^{1/x}\]

\[\to (+\infty)^{+\infty}=\infty\]

Answer 6

Wait, I'm having some doubts about that continuity sentence...

Answer 7

using the fact that \[\cos(-x)=\cos(x)\]

and \[1-\cos(x)>0\] near 0 (not equal to zero)

Answer 8

Yes, use @Zarkon's method. Mine doesn't work, since the natural log isn't defined for \(x<0\).

Answer 9

And thus continuity can't be used in this case.

Answer 10

but this limit is supposed to be done with L'hospital rule, not the other method

Answer 11

ahh...but i like my way ;)

Answer 12

1- cosx > 0 so the ln is not a problem...

Answer 13

As do I, much simpler. I suppose you can rewrite the way Zarkon did in his first post, then apply the logarithmic method. Continuity can be used since you'd be taking the limit as \(x\to0^+\) this time around.

Answer 14

i like it too @Zarkon

Answer 15

but ur method doesnt crash there
as ln (1-cosx) existis......1-cosx >0!!!!!

Answer 16

Oh right, I forgot all about the \(1-\cos x\) bit. Carry on :)

Answer 17

the problem is that u get to exp (sinx/(1-cosx))

Answer 18

after thar u get exp (cosx/sinx)

Answer 19

wich gives u the wrong asnwer

Answer 20

How is it wrong? Off by a sign or something?

Answer 21

yeah

Answer 22

when x goes to 0+ the limit is +inf and it should be 0
and when x goes to 0- the limit is 0 and should be + inf

Answer 23

\[\exp\Big(\lim_{x\to0^-}\frac{\cos x}{\sin x}\Big)~\Rightarrow~\exp\Big(\frac{1}{0}\Big)~\Rightarrow~\cdots\]

Answer 24

the 2 side limits are different

Answer 25

using what u said they would be equal

Answer 26

lim x to 0- of exp (cosx/sinx) = exp (-inf) = 0

Answer 27

\[\exp{\Bigg(\lim_{x\to0^-}\frac{1}{x}\ln(1-\cos x)}\Bigg)\]

you don't use L'Hospitals rule here

Answer 28

\[\lim_{x\to0^-}\frac{1}{x}=-\infty\]
\[\lim_{x\to0^-}\ln(1-\cos x)=-\infty\]

\[e^{-\infty\times-\infty}=e^{\infty}=\infty\]

Answer 29

^^ and there it is!

Answer 30

i know that but what I know also is that if I say that in an exam I'm getting a D-

Answer 31

u can't say exp (-inf*-inf) = exp (inf)
at least the last time I wrote that i got an X as correction!

Answer 32

Any prof that would give this problem and say use L'Hospitals rule is nuts.


Also I was abusing notation above just to make it so I didn't have to type as much

Answer 33

i hope ur right!

Answer 34

sorry to disturb u

Answer 35

I would never give this problem to any of my students and then force them to use L'Hospitals rule

Answer 36

help me with one more then? >.<

Answer 37

Zakron will u fan me?

Answer 38

why should I fan you?

Answer 39

what do i get out of this transaction ?

Answer 40

Aren't u a mod?

Answer 41

yes

Answer 42

that doesn't require me to "fan" other users

Answer 43

just for my own curiosity, why do you get to replace \(1-\cos(x)\) by \(\frac{1}{1-\cos(x)}\)?

Answer 44

\[\lim_{x \rightarrow 1-}\frac{ e ^{1/\left( x^2 -1\right)} }{ x-1 }\]

Answer 45

I changes the limit from \(x\to0^{-}\) to \(x\to0^{+}\)

Answer 46

Well u would help me if I could message u something

Answer 47

let u=-x

Answer 48

\[(\cdot)^{1/x}\]

becomes

\[(\cdot)^{-1/x}\]

Answer 49

Also aren't mods admins?

Answer 50

I dont see where u=-x could help me =(

Answer 51

oooh i didn't see the change of \(-\frac{1}{x}\) in the exponent!

Answer 52

Sat u an admin? And can u close an acct for me?

Answer 53

matheusoliveira...I was referring to the original problem

Answer 54

o sorry, but can u help in the last one please?

Answer 55

@selrachcw95 no i am not an administrator

Answer 56

Ok.

Answer 57

use L'Hospitals...

Answer 58

but then I get -2xe^(1/(x^2-1))/((x^2-1)^2)
right?

Answer 59

just to put in my two cents

\[(1-\cos(x))^{\frac{1}{x}}\] is by definition
\[e^{\frac{\ln(1-\cos(x))}{x}}\] take the limit in the exponent, for which l'hopital cannot be applied, and get \(\infty\) by inspection

Answer 60

i'm talking about
\[\lim_{x \rightarrow 1-}\frac{ e ^{1/\left( x^2 -1\right)} }{ x-1 }\]

Answer 61

answer please @Zarkon

Answer 62

try this let \[u=\frac{1}{x^2-1}\]

Answer 63

it works out...but I don't see a really 'clean' way to do it.

Answer 64

i can't make it work =X

Answer 65

\[\lim_{u \rightarrow -\infty} \frac{ e ^{u} }{ \sqrt{1/u + 1} -1}\]

Answer 66

right?

Answer 67

yes

Answer 68

you could also try

\[\lim_{x \to1^-}\frac{ e ^{1/( x^2 -1)} }{ x-1 }\]

\[\lim_{x \to1^-}-\frac{ e ^{1/( x^2 -1)} }{ 1-x }\]

\[\lim_{x \to1^-}- e ^{1/( x^2 -1)-\ln(1-x)} \]

Answer 69

\[\lim_{x\to 1^-}\left[\frac{1}{ x^2 -1}-\ln(1-x)\right]\]

Answer 70

now what do I do?

Answer 71

\[\lim_{x\to 1^-}\left[\frac{1-(x^2-1)\ln(1-x)}{ x^2 -1}\right]\]

Answer 72

do you see what this limit is?

Answer 73

no =X

Answer 74

not an easy limit ;)

Answer 75

it would be better to keep the ln(x-1) and keep the signal in the original equation i think

Answer 76

x-1 is negative

Answer 77

hum

Answer 78

that is why i used 1-x

Answer 79

so how do I progress?

Answer 80

this limit is in the L'hospital's rule section of the course =X

Answer 81

start by showing that \((x^2-1)\ln(1-x)\to 0\text{ as }x\to1^{-}\)

Answer 82

what would give us

\[\lim_{x\to 1^-}\left[\frac{1-(x^2-1)\ln(1-x)}{ x^2 -1}\right]=-\infty\]

and thus the overall limit (abusing notation) would be \[e^{-\infty}=0\]

Answer 83

(x2−1)ln(1−x)→0 as x→1- how so?

Answer 84

write as \[\frac{\ln(1-x)}{1/(x^2-1)}\] and use L'Hospitals rule

Answer 85

that gives me -1

Answer 86

\[\frac{-1/(1-x)}{-2x/(x^2-1)^2}\]
right?

Answer 87

yep

Answer 88

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

Answer 89

(x+1)(x-1)/2x(1-x) = -(x+1)/2x = -2/2 = -1

Answer 90

no

Answer 91

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

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

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

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

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

Answer 92

now let x=1

Answer 93

yeah its -inf ur right!!!!

Answer 94

my bad

Answer 95

thx a lot and sorry for my stupidity!!! =X