Question

Lim{1+(arc cosx)^1-x } when x tends to 1-

Answer 1

please show any work done so far

Answer 2

Lt (1+(arc cosx)^h)

Answer 3

I replaced x by 1-h

Answer 4

Sry (1+(arc cos(1-h))^h)

Answer 5

Is this the question
$$ \Large \lim_{x \to 1^{-}} 1 + (\arccos x)^{1-x} $$

Answer 6

Yaa correct

Answer 7

After this i m not getting d way out to solve this...

Answer 8

where does h go, as x -> 1-

Answer 9

We can write x as 1-h

Answer 10

where does h tend to as x approaches 1 from below (or from the left of 1)

Answer 11

h tends to 0

Answer 12

i have assumed that h is very minimal term as that we could replace x by 1-h

Answer 13

yes you can do that.

Answer 14

But after that how to move on with this

Answer 15

I don't see how we can move on at this point.
have you thought about using a logarithm

Answer 16

Well ,for the term (arc cos(1-h))^h we can use that

Answer 17

For instance, we use logs to find the limit of
$$ \large\lim_{x \to 0^{+}} x^x $$

Answer 18

But the biggest fact is that i really didn't think about it unless u have asked me to think on it

Answer 19

if you plug in h = 0 into (arc cos(1-h))^h
we get (arc cos(1-0))^0 = ( arcos 1 ) ^0 = 0 ^0
which is indeterminate form

Answer 20

In such forms only we use logarithm 0^0 or infinity ^ infinity

Answer 21

correct, thats a standard approach for
limit x->a f(x)^g(x) , as f(x) ->0 , g(x) ->0
or you can find the limit of e^ln(f(x)^g(x))

Answer 22

would you like to use logs on your expression?

Answer 23

So we would be left with
log y= h log arc cos(1-h)

Answer 24

I don't know how to proceed frm here

Answer 25

by taking logs on both sides,
we convert a 0^0 expression into 0*0 expression,
wherein we can now plug in the limiting value in the function :)

Answer 26

in an easy language,
now just put h=1 :P

Answer 27

What is P here?

Answer 28

Oh sorry i don't get it at frst

Answer 29

Bt h tends to 0 how can we put 1 now

Answer 30

sorry, typo then.
plug in whatever h approaches to :)

Answer 31

Log 0 is not defined then

Answer 32

as \(x \to 1^- \\ h \to 0^+\)

do you understand this?

Answer 33

Yup

Answer 34

So u want to say that 1-h is not exactly 1 it is sumthing 0.99999 sort of and its cos inverse will lead me to sumthin 0.00001 sort

Answer 35

yes, but still log of that is a very high negative number \(\to -\infty \)

Answer 36

Yup so what now....

Answer 37

U said that it would be 0

Answer 38

Guys what now........

Answer 39

you can apply Lhopitals rule to an expression 0 * (-infinity)

Answer 40

The formula can be used bt this would be complicated as again we r going to be left vid 0/0 form

Answer 41

$$ \lim_{h \to 0^{+}} ~ 1 + (\arccos(1-h))^h
\\ \ \\ = \lim_{h \to 0^{+}} ~ 1 + \lim_{h \to 0^{+}} (\arccos(1-h))^h
\\ \ \\ = 1 + \lim_{h \to 0^{+}} (\arccos(1-h))^h
$$

Answer 42

we only need to concentrate on finding the limit of \( \lim_{h \to 0^{+}} (\arccos(1-h))^h \)

Answer 43

Yaa correct

Answer 44

This is what i n @hartnn discussed...:)

Answer 45

I got upto this point

Answer 46

suppose that
$$ \lim_{h \to 0^{+}} (\arccos(1-h))^h = y
\\ \ \\ \text { then }
\\ \ \\ \ln (y) = \ln( \lim_{h \to 0^{+}} (\arccos(1-h))^h)
\\ \ \\ =\lim_{h \to 0^{+}} \ln (\arccos(1-h))^h
\\ \ \\ =\lim_{h \to 0^{+}} h \cdot \ln (\arccos(1-h))= 0 \cdot (-\infty)
\\ \ \\ =\lim_{h \to 0^{+}} \frac{ \ln (\arccos(1-h))}{1 / h }
\\ \ \\ \implies
$$

Answer 47

now you can use Lhopitals rule

Answer 48

Yaa right....this last step was also apt...bt when we wil apply l hospital then again 0/0 form is leftover

Answer 49

it shouldn't if you did it correctly

Answer 50

this is actually more difficult than it needs to be because you made that substitution earlier
it would be easier with the original x variable

Answer 51

That's what my professor used to tell me that it"s bettr to convert into h .This is why i started with all that...

Answer 52

not in every case is it better to convert to h
in some cases it does make it simpler, in others it does not

Answer 53

Yaa correct i have got dis...

Answer 54

Now when we will substitute h as 0 it would give us 0/0 form

Answer 55

lets simplify first

Answer 56

did you get this?
yes you are right , this is another indeterminate form.
we can apply Lhopital again, if that doesn't work, then drop it and try a different approach.

Answer 57

I told u already.....:)..anyways u r right we can try that

Answer 58

U put 1-(1-h)^2 as h^2 ...How?

Answer 59

thats a major typo

Answer 60

??

Answer 61

\[\text{if} ~~ \lim_{h \to 0^{+}} (\arccos(1-h))^h = y
\\ \ \\ \text { then }
\\ \ \\ \ln (y) = \ln( \lim_{h \to 0^{+}} (\arccos(1-h))^h)
\\ \ \\ =\lim_{h \to 0^{+}} \ln (\arccos(1-h))^h
\\ \ \\ =\lim_{h \to 0^{+}} h \cdot \ln (\arccos(1-h))= 0 \cdot (-\infty)
\\ \ \\ =\lim_{h \to 0^{+}} \frac{ \ln (\arccos(1-h))}{1 / h }
\\ \ \\\Large \implies
\lim_{h \to 0^{+}} \frac{ \frac{1}{\arccos(1-h) } \cdot \frac{-1}{\sqrt{ 1- (1-h)^2} } \cdot (-1)}{-1 / h^2 }
\\ \ \\
\Large \lim_{h \to 0^{+}} \frac{ \frac{1}{\arccos(1-h) } \cdot \frac{-1}{\sqrt{ 1- (1 - 2h + h^2)} } \cdot (-1)}{-1 / h^2 }




\]

Answer 62

Yes now its fine

Answer 63

Now actually it would be h^2 whole upon arc cos(1-h)√2h-h^2

Answer 64

$$\Large \implies
\\ \ \\ \Large =
\lim_{h \to 0^{+}} \frac{ \frac{1}{\arccos(1-h) } \cdot \frac{-1}{\sqrt{ 1- (1-h)^2} } \cdot (-1)}{-1 / h^2 }
\\ \ \\
\Large = \lim_{h \to 0^{+}} \frac{ \frac{1}{\arccos(1-h) } \cdot \frac{-1}{\sqrt{ 1-1 + 2h -h^2)} } \cdot (-1)}{-1 / h^2 }
\\ \ \\
\Large = \lim_{h \to 0^{+}} \frac{ \frac{1}{\arccos(1-h) } \cdot \frac{1}{\sqrt{ 2h -h^2} }}{-1 / h^2 }
\\ \ \\
\Large = \lim_{h \to 0^{+}} \frac{ \frac{1}{\arccos(1-h) } \cdot \frac{1}{\sqrt{ h( 2 -h)} }}{-1 / h^2 }



$$

Answer 65

We can take h^2 common from root and we will be left with h√2/h-1

Answer 66

\[\Large \implies
\\ \ \\ \Large =
\lim_{h \to 0^{+}} \frac{ \frac{1}{\arccos(1-h) } \cdot \frac{-1}{\sqrt{ 1- (1-h)^2} } \cdot (-1)}{-1 / h^2 }
\\ \ \\
\Large = \lim_{h \to 0^{+}} \frac{ \frac{1}{\arccos(1-h) } \cdot \frac{-1}{\sqrt{ 1-1 + 2h -h^2)} } \cdot (-1)}{-1 / h^2 }
\\ \ \\
\Large = \lim_{h \to 0^{+}} \frac{ \frac{1}{\arccos(1-h) } \cdot \frac{1}{\sqrt{ 2h -h^2} }}{-1 / h^2 }
\\ \ \\
\Large = \lim_{h \to 0^{+}} \frac{ \frac{1}{\arccos(1-h) } \cdot \frac{1}{\sqrt{ h( 2 -h)} }}{-1 / h^2 }
\\ \ \\
\Large = \lim_{h \to 0^{+}} \frac{1}{\arccos(1-h) } \cdot \frac{1}{\sqrt{ h( 2 -h)} } \cdot \frac{-h^2}{ 1 }



\]

Answer 67

Hey...hey....it's coming out to be 1 if we cancel h^2 with the h in denominator n then 2/h would be infinite and this term is in denominator we will get it to be 0 now log 1 is 0 so its limit is 1

Answer 68

\[\Large \implies
\\ \ \\ \Large =
\lim_{h \to 0^{+}} \frac{ \frac{1}{\arccos(1-h) } \cdot \frac{-1}{\sqrt{ 1- (1-h)^2} } \cdot (-1)}{-1 / h^2 }
\\ \ \\
\Large = \lim_{h \to 0^{+}} \frac{ \frac{1}{\arccos(1-h) } \cdot \frac{-1}{\sqrt{ 1-1 + 2h -h^2)} } \cdot (-1)}{-1 / h^2 }
\\ \ \\
\Large = \lim_{h \to 0^{+}} \frac{ \frac{1}{\arccos(1-h) } \cdot \frac{1}{\sqrt{ 2h -h^2} }}{-1 / h^2 }
\\ \ \\
\Large = \lim_{h \to 0^{+}} \frac{ \frac{1}{\arccos(1-h) } \cdot \frac{1}{\sqrt{ h( 2 -h)} }}{-1 / h^2 }
\\ \ \\
\Large = \lim_{h \to 0^{+}} \frac{1}{\arccos(1-h) } \cdot \frac{1}{\sqrt{ h} \sqrt{ 2 -h} } \cdot \frac{-h^2}{ 1 }
\\ \ \\
\Large = \lim_{h \to 0^{+}} \frac{-h^{3/2}}{\arccos(1-h) \sqrt{2-h} }


\]

Answer 69

can you draw that out?

Answer 70

I m not getting how to draw it out....

Answer 71

The draw tool?

Answer 72

But i m not sure if we can take h2 common in the root or not...

Answer 73

Bcz my tchr has tolde we can do so when limit tends to infinite then we take out terms with higher power common n solve the limit thereafter...

Answer 74

What say @hartnn ?

Answer 75

the limit should come out to be zero
since ln(y) = 0 implies y = e^0 = 1

Answer 76

Bt am i doing it right i mean can we take common like that...i m confused over it

Answer 77

I don't see how you can do that. can you explain it again

Answer 78

If we go back a few steps, we can try to introduce 1/h^2 inside the radical.

Answer 79

Yes you try to do that take h^2 common in the radical itself and solve it

Answer 80

\[\Large \implies
\\ \ \\ \Large =
\lim_{h \to 0^{+}} \frac{ \frac{1}{\arccos(1-h) } \cdot \frac{-1}{\sqrt{ 1- (1-h)^2} } \cdot (-1)}{-1 / h^2 }
\\ \ \\
\Large = \lim_{h \to 0^{+}} \frac{ \frac{1}{\arccos(1-h) } \cdot \frac{-1}{\sqrt{ 1-1 + 2h -h^2)} } \cdot (-1)}{-1 / h^2 }
\\ \ \\
\Large = \lim_{h \to 0^{+}} \frac{ -h^2 }{\arccos(1-h) {\sqrt{ 2h -h^2} }}
\\ \ \\

\]

Answer 81

that would still leave an h in the numerator

Answer 82

I don't think the substitution made was a productive substitution.

Answer 83

There are two things you can do
1. go back and use the original x, dont make substitution
2. make a linear (or quadratic) approximation for arccos(x) near x = 1

Answer 84

Linear approximation in what sense ....

Answer 85

for example if you want to find
$$\large \lim_{x \to 0^{+}} (\sin x )^ x \approx \lim_{x \to 0^{+}} (x )^ x $$ you can approximate sin x by x

Answer 86

y=sin x behaves like y=x , as x approaches zero

Answer 87

Bt in this particular case what sort of approximation are you gonna use...

Answer 88

what function does arccos x behave like, as x approaches 1

Answer 89

im going to try this problem using the original variable x instead of h

Answer 90

Actually u r the first person who has told me about such a method so i really don't know such a function....

Answer 91

:) i m really sorry but i m taking a whole lot of ur precious time in this one question

Answer 92

Alryt u do it ur way ....

Answer 93

\[ \Large \lim_{x \to 1^{-}} \ln(\arccos x) ^{1-x} = \lim_{x \to 1^{-}} (1-x) \cdot \ln(\arccos x)
\\ \ \\ \Large = \lim_{x \to 1^{-}} \frac{\ln(\arccos x) }{\frac{1}{1-x}}
\\ \ \\ \Large \implies \lim_{x \to 1^{-}} \frac{ \frac{ -1 }{\arccos x~ \sqrt{1-x^2} } } { \frac{1}{(1-x)^2} }


\]

Answer 94

This is also 0/0 form

Answer 95

would it be easier if we bring arccos in the denominator?

(1-x) divided by 1/ln (arccos x)

Answer 96

\[ \large \lim_{x \to 1^{-}} \ln(\arccos x) ^{1-x} = \lim_{x \to 1^{-}} (1-x) \cdot \ln(\arccos x)
\\ \ \\ \large = \lim_{x \to 1^{-}} \frac{\ln(\arccos x) }{\frac{1}{1-x}}
\\ \ \\ \large \implies \lim_{x \to 1^{-}} \frac{ \frac{ -1 }{\arccos x~ \sqrt{1-x^2} } } { \frac{1}{(1-x)^2} }
= \large \lim_{x \to 1^{-}} \frac{ -1 }{\arccos x~ \sqrt{1-x^2} } \cdot \frac{(1-x)^2}{1}


\]

Answer 97

hartnn
it might be, didn't try that :)

Answer 98

Isn't this one question just too much for each of us...