How do you evaluate limit of $fracsqrt1+x^2 - sqrt 1+xsqrt 1+x^3 - sqrt 1+x$ when $x$ tends to $0$?

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I tried rationalization method and got as $fracx^2-xsqrt 1+x^3 - sqrt 1+x (sqrt1+x^2 + sqrt 1+x)$. But i feel the denominator having power of 3 I may be doing it wrong.The answer should be 1. Please help.

calculus limits

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edited Jul 15 at 9:57

asked Jul 15 at 8:25

Shaikh Sakib

184

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Use Binomial expansion for each term
  – markovchain
  Jul 15 at 8:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I split the denominator's 1+x^3 into (1+x)(x^2-x+1) but how should i multiply it with root of (1+x^2) + root of (1+x). Getting confused there.
  – Shaikh Sakib
  Jul 15 at 8:48
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $(1+x)^1/2 =1+x/2+(1/2)(1-1/2)x^2+...$ use this expression for all terms and then take limit to zero, that will be very easy.
  – markovchain
  Jul 15 at 8:53
  
  

  
  
  
  

add a comment | 

up vote
3
down vote

favorite

I tried rationalization method and got as $fracx^2-xsqrt 1+x^3 - sqrt 1+x (sqrt1+x^2 + sqrt 1+x)$. But i feel the denominator having power of 3 I may be doing it wrong.The answer should be 1. Please help.

calculus limits

share|cite|improve this question

edited Jul 15 at 9:57

asked Jul 15 at 8:25

Shaikh Sakib

184

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Use Binomial expansion for each term
  – markovchain
  Jul 15 at 8:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I split the denominator's 1+x^3 into (1+x)(x^2-x+1) but how should i multiply it with root of (1+x^2) + root of (1+x). Getting confused there.
  – Shaikh Sakib
  Jul 15 at 8:48
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $(1+x)^1/2 =1+x/2+(1/2)(1-1/2)x^2+...$ use this expression for all terms and then take limit to zero, that will be very easy.
  – markovchain
  Jul 15 at 8:53
  
  

  
  
  
  

add a comment | 

up vote
3
down vote

favorite

up vote
3
down vote

favorite

I tried rationalization method and got as $fracx^2-xsqrt 1+x^3 - sqrt 1+x (sqrt1+x^2 + sqrt 1+x)$. But i feel the denominator having power of 3 I may be doing it wrong.The answer should be 1. Please help.

calculus limits

share|cite|improve this question

edited Jul 15 at 9:57

asked Jul 15 at 8:25

Shaikh Sakib

184

I tried rationalization method and got as $fracx^2-xsqrt 1+x^3 - sqrt 1+x (sqrt1+x^2 + sqrt 1+x)$. But i feel the denominator having power of 3 I may be doing it wrong.The answer should be 1. Please help.

calculus limits

share|cite|improve this question

edited Jul 15 at 9:57

asked Jul 15 at 8:25

Shaikh Sakib

184

share|cite|improve this question

share|cite|improve this question

edited Jul 15 at 9:57

edited Jul 15 at 9:57

edited Jul 15 at 9:57

asked Jul 15 at 8:25

Shaikh Sakib

184

asked Jul 15 at 8:25

Shaikh Sakib

184

asked Jul 15 at 8:25

Shaikh Sakib

184

184

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Use Binomial expansion for each term
  – markovchain
  Jul 15 at 8:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I split the denominator's 1+x^3 into (1+x)(x^2-x+1) but how should i multiply it with root of (1+x^2) + root of (1+x). Getting confused there.
  – Shaikh Sakib
  Jul 15 at 8:48
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $(1+x)^1/2 =1+x/2+(1/2)(1-1/2)x^2+...$ use this expression for all terms and then take limit to zero, that will be very easy.
  – markovchain
  Jul 15 at 8:53
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Use Binomial expansion for each term
  – markovchain
  Jul 15 at 8:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I split the denominator's 1+x^3 into (1+x)(x^2-x+1) but how should i multiply it with root of (1+x^2) + root of (1+x). Getting confused there.
  – Shaikh Sakib
  Jul 15 at 8:48
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $(1+x)^1/2 =1+x/2+(1/2)(1-1/2)x^2+...$ use this expression for all terms and then take limit to zero, that will be very easy.
  – markovchain
  Jul 15 at 8:53
  
  

  
  
  
  

Use Binomial expansion for each term
– markovchain
Jul 15 at 8:31

Use Binomial expansion for each term
– markovchain
Jul 15 at 8:31

I split the denominator's 1+x^3 into (1+x)(x^2-x+1) but how should i multiply it with root of (1+x^2) + root of (1+x). Getting confused there.
– Shaikh Sakib
Jul 15 at 8:48

I split the denominator's 1+x^3 into (1+x)(x^2-x+1) but how should i multiply it with root of (1+x^2) + root of (1+x). Getting confused there.
– Shaikh Sakib
Jul 15 at 8:48

$(1+x)^1/2 =1+x/2+(1/2)(1-1/2)x^2+...$ use this expression for all terms and then take limit to zero, that will be very easy.
– markovchain
Jul 15 at 8:53

$(1+x)^1/2 =1+x/2+(1/2)(1-1/2)x^2+...$ use this expression for all terms and then take limit to zero, that will be very easy.
– markovchain
Jul 15 at 8:53

add a comment | 

5 Answers
5

active

oldest

votes

up vote
2
down vote



accepted

Use the fact that when $uto 0$, $(1+u)^n approx 1+nu$.

Using that fact,

$$beginalign L &= lim_xto 0 fracsqrt1+x^2 - sqrt 1+xsqrt 1+x^3 - sqrt 1+x \ &= lim_xto 0 dfrac1+dfrac 12 x^2 - (1+dfrac 12 x)1+dfrac 12 x^3 - (1+dfrac 12 x)\ &= lim_xto 0 dfracx^2-xx^3-x \ &= lim_xto 0dfrac11+x \ &= colorred1endalign$$

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answered Jul 15 at 10:07

DarkKnight

403210

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  1
  

  
  
  
  
  
  

  
  the most hassle free and straight forward answer. Thank you.
  – Shaikh Sakib
  Jul 15 at 11:03
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If you like the answer, you can accept it by clicking on the tick below the upvote button.
  – DarkKnight
  Jul 15 at 14:00
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  upvoted already!
  – Shaikh Sakib
  Jul 15 at 15:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Shaikh Sakib math.meta.stackexchange.com/a/3287/537100
  – DarkKnight
  Jul 15 at 16:17
  

  
  
  
  

add a comment | 

up vote
3
down vote

Using the same rationalization method, when $xnotin0,1$,
$$
fracsqrt1+x^2-sqrt1+xsqrt1+x^3-sqrt1+x
=frac1x+1fracsqrt1+x^3+sqrt1+xsqrt1+x^2+sqrt1+x
$$

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edited Jul 15 at 8:46

answered Jul 15 at 8:32

robjohn♦

258k26297612

add a comment | 

up vote
2
down vote

Use definition of derivation:
beginalign
fracsqrt1+x^2 - sqrt 1+xsqrt 1+x^3 - sqrt 1+x
&=
fracfracsqrt1+x^2-1x - fracsqrt 1+x-1xfracsqrt 1+x^3-1x - fracsqrt 1+x-1x\
&=
frac(sqrt1+x^2)'(sqrt 1+x^3)'\
&=
dfrac0-frac120-frac12\
&=1
endalign

share|cite|improve this answer

edited Jul 15 at 8:50

answered Jul 15 at 8:48

Nosrati

19.9k41644

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You have miscomputed the derivatives of a couple of the terms.
  – robjohn♦
  Jul 15 at 8:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @robjohn fixed. thanks!
  – Nosrati
  Jul 15 at 8:52
  

  
  
  
  

add a comment | 

up vote
1
down vote

$sqrt1+x^n=1+x^n/2+O(x^2n)$ as $xto0$.
In particular $sqrt1+x=1+x/2+O(x^2)$, $sqrt1+x^2=1+O(x^2)$
and $sqrt1+x^3=1+O(x^2)$. Then your fraction is
$$frac1+O(x^2)-(1+x/2+O(x^2))1+O(x^2)-(1+x/2+O(x^2))
=frac-x/2+O(x^2)-x/2+O(x^2)$$
etc.

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answered Jul 15 at 8:32

Lord Shark the Unknown

85.8k951112

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up vote
0
down vote

You should rationalize both numerator and denominator.
Terms will get cancelled and you will get the final answer

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answered Jul 15 at 17:16

Bhanupratapsingh Sengar

11

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Use the fact that when $uto 0$, $(1+u)^n approx 1+nu$.

Using that fact,

$$beginalign L &= lim_xto 0 fracsqrt1+x^2 - sqrt 1+xsqrt 1+x^3 - sqrt 1+x \ &= lim_xto 0 dfrac1+dfrac 12 x^2 - (1+dfrac 12 x)1+dfrac 12 x^3 - (1+dfrac 12 x)\ &= lim_xto 0 dfracx^2-xx^3-x \ &= lim_xto 0dfrac11+x \ &= colorred1endalign$$

share|cite|improve this answer

answered Jul 15 at 10:07

DarkKnight

403210

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  the most hassle free and straight forward answer. Thank you.
  – Shaikh Sakib
  Jul 15 at 11:03
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If you like the answer, you can accept it by clicking on the tick below the upvote button.
  – DarkKnight
  Jul 15 at 14:00
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  upvoted already!
  – Shaikh Sakib
  Jul 15 at 15:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Shaikh Sakib math.meta.stackexchange.com/a/3287/537100
  – DarkKnight
  Jul 15 at 16:17
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

Use the fact that when $uto 0$, $(1+u)^n approx 1+nu$.

Using that fact,

$$beginalign L &= lim_xto 0 fracsqrt1+x^2 - sqrt 1+xsqrt 1+x^3 - sqrt 1+x \ &= lim_xto 0 dfrac1+dfrac 12 x^2 - (1+dfrac 12 x)1+dfrac 12 x^3 - (1+dfrac 12 x)\ &= lim_xto 0 dfracx^2-xx^3-x \ &= lim_xto 0dfrac11+x \ &= colorred1endalign$$

share|cite|improve this answer

answered Jul 15 at 10:07

DarkKnight

403210

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  the most hassle free and straight forward answer. Thank you.
  – Shaikh Sakib
  Jul 15 at 11:03
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If you like the answer, you can accept it by clicking on the tick below the upvote button.
  – DarkKnight
  Jul 15 at 14:00
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  upvoted already!
  – Shaikh Sakib
  Jul 15 at 15:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Shaikh Sakib math.meta.stackexchange.com/a/3287/537100
  – DarkKnight
  Jul 15 at 16:17
  

  
  
  
  

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Use the fact that when $uto 0$, $(1+u)^n approx 1+nu$.

Using that fact,

$$beginalign L &= lim_xto 0 fracsqrt1+x^2 - sqrt 1+xsqrt 1+x^3 - sqrt 1+x \ &= lim_xto 0 dfrac1+dfrac 12 x^2 - (1+dfrac 12 x)1+dfrac 12 x^3 - (1+dfrac 12 x)\ &= lim_xto 0 dfracx^2-xx^3-x \ &= lim_xto 0dfrac11+x \ &= colorred1endalign$$

share|cite|improve this answer

answered Jul 15 at 10:07

DarkKnight

403210

Use the fact that when $uto 0$, $(1+u)^n approx 1+nu$.

Using that fact,

$$beginalign L &= lim_xto 0 fracsqrt1+x^2 - sqrt 1+xsqrt 1+x^3 - sqrt 1+x \ &= lim_xto 0 dfrac1+dfrac 12 x^2 - (1+dfrac 12 x)1+dfrac 12 x^3 - (1+dfrac 12 x)\ &= lim_xto 0 dfracx^2-xx^3-x \ &= lim_xto 0dfrac11+x \ &= colorred1endalign$$

share|cite|improve this answer

answered Jul 15 at 10:07

DarkKnight

403210

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 10:07

DarkKnight

403210

answered Jul 15 at 10:07

DarkKnight

403210

answered Jul 15 at 10:07

DarkKnight

403210

403210

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  the most hassle free and straight forward answer. Thank you.
  – Shaikh Sakib
  Jul 15 at 11:03
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If you like the answer, you can accept it by clicking on the tick below the upvote button.
  – DarkKnight
  Jul 15 at 14:00
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  upvoted already!
  – Shaikh Sakib
  Jul 15 at 15:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Shaikh Sakib math.meta.stackexchange.com/a/3287/537100
  – DarkKnight
  Jul 15 at 16:17
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  the most hassle free and straight forward answer. Thank you.
  – Shaikh Sakib
  Jul 15 at 11:03
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  If you like the answer, you can accept it by clicking on the tick below the upvote button.
  – DarkKnight
  Jul 15 at 14:00
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  upvoted already!
  – Shaikh Sakib
  Jul 15 at 15:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Shaikh Sakib math.meta.stackexchange.com/a/3287/537100
  – DarkKnight
  Jul 15 at 16:17
  

  
  
  
  

1

1

the most hassle free and straight forward answer. Thank you.
– Shaikh Sakib
Jul 15 at 11:03

the most hassle free and straight forward answer. Thank you.
– Shaikh Sakib
Jul 15 at 11:03

1

1

If you like the answer, you can accept it by clicking on the tick below the upvote button.
– DarkKnight
Jul 15 at 14:00

If you like the answer, you can accept it by clicking on the tick below the upvote button.
– DarkKnight
Jul 15 at 14:00

1

1

upvoted already!
– Shaikh Sakib
Jul 15 at 15:30

upvoted already!
– Shaikh Sakib
Jul 15 at 15:30

@Shaikh Sakib math.meta.stackexchange.com/a/3287/537100
– DarkKnight
Jul 15 at 16:17

@Shaikh Sakib math.meta.stackexchange.com/a/3287/537100
– DarkKnight
Jul 15 at 16:17

add a comment | 

up vote
3
down vote

Using the same rationalization method, when $xnotin0,1$,
$$
fracsqrt1+x^2-sqrt1+xsqrt1+x^3-sqrt1+x
=frac1x+1fracsqrt1+x^3+sqrt1+xsqrt1+x^2+sqrt1+x
$$

share|cite|improve this answer

edited Jul 15 at 8:46

answered Jul 15 at 8:32

robjohn♦

258k26297612

add a comment | 

up vote
3
down vote

Using the same rationalization method, when $xnotin0,1$,
$$
fracsqrt1+x^2-sqrt1+xsqrt1+x^3-sqrt1+x
=frac1x+1fracsqrt1+x^3+sqrt1+xsqrt1+x^2+sqrt1+x
$$

share|cite|improve this answer

edited Jul 15 at 8:46

answered Jul 15 at 8:32

robjohn♦

258k26297612

add a comment | 

up vote
3
down vote

up vote
3
down vote

Using the same rationalization method, when $xnotin0,1$,
$$
fracsqrt1+x^2-sqrt1+xsqrt1+x^3-sqrt1+x
=frac1x+1fracsqrt1+x^3+sqrt1+xsqrt1+x^2+sqrt1+x
$$

share|cite|improve this answer

edited Jul 15 at 8:46

answered Jul 15 at 8:32

robjohn♦

258k26297612

Using the same rationalization method, when $xnotin0,1$,
$$
fracsqrt1+x^2-sqrt1+xsqrt1+x^3-sqrt1+x
=frac1x+1fracsqrt1+x^3+sqrt1+xsqrt1+x^2+sqrt1+x
$$

share|cite|improve this answer

edited Jul 15 at 8:46

answered Jul 15 at 8:32

robjohn♦

258k26297612

share|cite|improve this answer

share|cite|improve this answer

edited Jul 15 at 8:46

edited Jul 15 at 8:46

edited Jul 15 at 8:46

answered Jul 15 at 8:32

robjohn♦

258k26297612

answered Jul 15 at 8:32

robjohn♦

258k26297612

answered Jul 15 at 8:32

robjohn♦

258k26297612

258k26297612

add a comment | 

add a comment | 

up vote
2
down vote

Use definition of derivation:
beginalign
fracsqrt1+x^2 - sqrt 1+xsqrt 1+x^3 - sqrt 1+x
&=
fracfracsqrt1+x^2-1x - fracsqrt 1+x-1xfracsqrt 1+x^3-1x - fracsqrt 1+x-1x\
&=
frac(sqrt1+x^2)'(sqrt 1+x^3)'\
&=
dfrac0-frac120-frac12\
&=1
endalign

share|cite|improve this answer

edited Jul 15 at 8:50

answered Jul 15 at 8:48

Nosrati

19.9k41644

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You have miscomputed the derivatives of a couple of the terms.
  – robjohn♦
  Jul 15 at 8:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @robjohn fixed. thanks!
  – Nosrati
  Jul 15 at 8:52
  

  
  
  
  

add a comment | 

up vote
2
down vote

Use definition of derivation:
beginalign
fracsqrt1+x^2 - sqrt 1+xsqrt 1+x^3 - sqrt 1+x
&=
fracfracsqrt1+x^2-1x - fracsqrt 1+x-1xfracsqrt 1+x^3-1x - fracsqrt 1+x-1x\
&=
frac(sqrt1+x^2)'(sqrt 1+x^3)'\
&=
dfrac0-frac120-frac12\
&=1
endalign

share|cite|improve this answer

edited Jul 15 at 8:50

answered Jul 15 at 8:48

Nosrati

19.9k41644

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You have miscomputed the derivatives of a couple of the terms.
  – robjohn♦
  Jul 15 at 8:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @robjohn fixed. thanks!
  – Nosrati
  Jul 15 at 8:52
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

Use definition of derivation:
beginalign
fracsqrt1+x^2 - sqrt 1+xsqrt 1+x^3 - sqrt 1+x
&=
fracfracsqrt1+x^2-1x - fracsqrt 1+x-1xfracsqrt 1+x^3-1x - fracsqrt 1+x-1x\
&=
frac(sqrt1+x^2)'(sqrt 1+x^3)'\
&=
dfrac0-frac120-frac12\
&=1
endalign

share|cite|improve this answer

edited Jul 15 at 8:50

answered Jul 15 at 8:48

Nosrati

19.9k41644

Use definition of derivation:
beginalign
fracsqrt1+x^2 - sqrt 1+xsqrt 1+x^3 - sqrt 1+x
&=
fracfracsqrt1+x^2-1x - fracsqrt 1+x-1xfracsqrt 1+x^3-1x - fracsqrt 1+x-1x\
&=
frac(sqrt1+x^2)'(sqrt 1+x^3)'\
&=
dfrac0-frac120-frac12\
&=1
endalign

share|cite|improve this answer

edited Jul 15 at 8:50

answered Jul 15 at 8:48

Nosrati

19.9k41644

share|cite|improve this answer

share|cite|improve this answer

edited Jul 15 at 8:50

edited Jul 15 at 8:50

edited Jul 15 at 8:50

answered Jul 15 at 8:48

Nosrati

19.9k41644

answered Jul 15 at 8:48

Nosrati

19.9k41644

answered Jul 15 at 8:48

Nosrati

19.9k41644

19.9k41644

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  You have miscomputed the derivatives of a couple of the terms.
  – robjohn♦
  Jul 15 at 8:50
  

  
  
  
  


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  @robjohn fixed. thanks!
  – Nosrati
  Jul 15 at 8:52
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You have miscomputed the derivatives of a couple of the terms.
  – robjohn♦
  Jul 15 at 8:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @robjohn fixed. thanks!
  – Nosrati
  Jul 15 at 8:52
  

  
  
  
  

You have miscomputed the derivatives of a couple of the terms.
– robjohn♦
Jul 15 at 8:50

You have miscomputed the derivatives of a couple of the terms.
– robjohn♦
Jul 15 at 8:50

@robjohn fixed. thanks!
– Nosrati
Jul 15 at 8:52

@robjohn fixed. thanks!
– Nosrati
Jul 15 at 8:52

add a comment | 

up vote
1
down vote

$sqrt1+x^n=1+x^n/2+O(x^2n)$ as $xto0$.
In particular $sqrt1+x=1+x/2+O(x^2)$, $sqrt1+x^2=1+O(x^2)$
and $sqrt1+x^3=1+O(x^2)$. Then your fraction is
$$frac1+O(x^2)-(1+x/2+O(x^2))1+O(x^2)-(1+x/2+O(x^2))
=frac-x/2+O(x^2)-x/2+O(x^2)$$
etc.

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answered Jul 15 at 8:32

Lord Shark the Unknown

85.8k951112

add a comment | 

up vote
1
down vote

$sqrt1+x^n=1+x^n/2+O(x^2n)$ as $xto0$.
In particular $sqrt1+x=1+x/2+O(x^2)$, $sqrt1+x^2=1+O(x^2)$
and $sqrt1+x^3=1+O(x^2)$. Then your fraction is
$$frac1+O(x^2)-(1+x/2+O(x^2))1+O(x^2)-(1+x/2+O(x^2))
=frac-x/2+O(x^2)-x/2+O(x^2)$$
etc.

share|cite|improve this answer

answered Jul 15 at 8:32

Lord Shark the Unknown

85.8k951112

add a comment | 

up vote
1
down vote

up vote
1
down vote

$sqrt1+x^n=1+x^n/2+O(x^2n)$ as $xto0$.
In particular $sqrt1+x=1+x/2+O(x^2)$, $sqrt1+x^2=1+O(x^2)$
and $sqrt1+x^3=1+O(x^2)$. Then your fraction is
$$frac1+O(x^2)-(1+x/2+O(x^2))1+O(x^2)-(1+x/2+O(x^2))
=frac-x/2+O(x^2)-x/2+O(x^2)$$
etc.

share|cite|improve this answer

answered Jul 15 at 8:32

Lord Shark the Unknown

85.8k951112

$sqrt1+x^n=1+x^n/2+O(x^2n)$ as $xto0$.
In particular $sqrt1+x=1+x/2+O(x^2)$, $sqrt1+x^2=1+O(x^2)$
and $sqrt1+x^3=1+O(x^2)$. Then your fraction is
$$frac1+O(x^2)-(1+x/2+O(x^2))1+O(x^2)-(1+x/2+O(x^2))
=frac-x/2+O(x^2)-x/2+O(x^2)$$
etc.

share|cite|improve this answer

answered Jul 15 at 8:32

Lord Shark the Unknown

85.8k951112

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 8:32

Lord Shark the Unknown

85.8k951112

answered Jul 15 at 8:32

Lord Shark the Unknown

85.8k951112

answered Jul 15 at 8:32

Lord Shark the Unknown

85.8k951112

85.8k951112

add a comment | 

add a comment | 

up vote
0
down vote

You should rationalize both numerator and denominator.
Terms will get cancelled and you will get the final answer

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answered Jul 15 at 17:16

Bhanupratapsingh Sengar

11

add a comment | 

up vote
0
down vote

You should rationalize both numerator and denominator.
Terms will get cancelled and you will get the final answer

share|cite|improve this answer

answered Jul 15 at 17:16

Bhanupratapsingh Sengar

11

add a comment | 

up vote
0
down vote

up vote
0
down vote

You should rationalize both numerator and denominator.
Terms will get cancelled and you will get the final answer

share|cite|improve this answer

answered Jul 15 at 17:16

Bhanupratapsingh Sengar

11

You should rationalize both numerator and denominator.
Terms will get cancelled and you will get the final answer

share|cite|improve this answer

answered Jul 15 at 17:16

Bhanupratapsingh Sengar

11

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 17:16

Bhanupratapsingh Sengar

11

answered Jul 15 at 17:16

Bhanupratapsingh Sengar

11

answered Jul 15 at 17:16

Bhanupratapsingh Sengar

11

11

add a comment | 

add a comment | 

 

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