Proving a limit for a fraction

Clash Royale CLAN TAG#URR8PPP

up vote
2
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Let $rin (0,1)$. How can I prove

$$lim_epsilonrightarrow 0frac(1+epsilon)^r+(1-epsilon)^r-2epsilon^2r=0?$$

limits

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edited Jul 19 at 16:00

Clayton

17.9k22882

asked Jul 19 at 15:59

Dai Jinaid

194

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since you're taking a limit as $varepsilonto0$, you don't "choose" $varepsilon$.
  – Clayton
  Jul 19 at 16:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are right. I am sorry. I wanted to fix r, not epsilon.
  – Dai Jinaid
  Jul 19 at 16:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Have you made any attempts on the problem?
  – Clayton
  Jul 19 at 16:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Try l'Hopital's rule
  – Green.H
  Jul 19 at 16:03
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Yes, l'Hopital doesn't help
  – Dai Jinaid
  Jul 19 at 16:10
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

Let $rin (0,1)$. How can I prove

$$lim_epsilonrightarrow 0frac(1+epsilon)^r+(1-epsilon)^r-2epsilon^2r=0?$$

limits

share|cite|improve this question

edited Jul 19 at 16:00

Clayton

17.9k22882

asked Jul 19 at 15:59

Dai Jinaid

194

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since you're taking a limit as $varepsilonto0$, you don't "choose" $varepsilon$.
  – Clayton
  Jul 19 at 16:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are right. I am sorry. I wanted to fix r, not epsilon.
  – Dai Jinaid
  Jul 19 at 16:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Have you made any attempts on the problem?
  – Clayton
  Jul 19 at 16:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Try l'Hopital's rule
  – Green.H
  Jul 19 at 16:03
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Yes, l'Hopital doesn't help
  – Dai Jinaid
  Jul 19 at 16:10
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Let $rin (0,1)$. How can I prove

$$lim_epsilonrightarrow 0frac(1+epsilon)^r+(1-epsilon)^r-2epsilon^2r=0?$$

limits

share|cite|improve this question

edited Jul 19 at 16:00

Clayton

17.9k22882

asked Jul 19 at 15:59

Dai Jinaid

194

Let $rin (0,1)$. How can I prove

$$lim_epsilonrightarrow 0frac(1+epsilon)^r+(1-epsilon)^r-2epsilon^2r=0?$$

limits

share|cite|improve this question

edited Jul 19 at 16:00

Clayton

17.9k22882

asked Jul 19 at 15:59

Dai Jinaid

194

share|cite|improve this question

share|cite|improve this question

edited Jul 19 at 16:00

Clayton

17.9k22882

edited Jul 19 at 16:00

Clayton

17.9k22882

edited Jul 19 at 16:00

Clayton

17.9k22882

17.9k22882

asked Jul 19 at 15:59

Dai Jinaid

194

asked Jul 19 at 15:59

Dai Jinaid

194

asked Jul 19 at 15:59

Dai Jinaid

194

194

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since you're taking a limit as $varepsilonto0$, you don't "choose" $varepsilon$.
  – Clayton
  Jul 19 at 16:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are right. I am sorry. I wanted to fix r, not epsilon.
  – Dai Jinaid
  Jul 19 at 16:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Have you made any attempts on the problem?
  – Clayton
  Jul 19 at 16:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Try l'Hopital's rule
  – Green.H
  Jul 19 at 16:03
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Yes, l'Hopital doesn't help
  – Dai Jinaid
  Jul 19 at 16:10
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since you're taking a limit as $varepsilonto0$, you don't "choose" $varepsilon$.
  – Clayton
  Jul 19 at 16:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are right. I am sorry. I wanted to fix r, not epsilon.
  – Dai Jinaid
  Jul 19 at 16:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Have you made any attempts on the problem?
  – Clayton
  Jul 19 at 16:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Try l'Hopital's rule
  – Green.H
  Jul 19 at 16:03
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Yes, l'Hopital doesn't help
  – Dai Jinaid
  Jul 19 at 16:10
  
  

  
  
  
  

Since you're taking a limit as $varepsilonto0$, you don't "choose" $varepsilon$.
– Clayton
Jul 19 at 16:01

Since you're taking a limit as $varepsilonto0$, you don't "choose" $varepsilon$.
– Clayton
Jul 19 at 16:01

You are right. I am sorry. I wanted to fix r, not epsilon.
– Dai Jinaid
Jul 19 at 16:03

You are right. I am sorry. I wanted to fix r, not epsilon.
– Dai Jinaid
Jul 19 at 16:03

Have you made any attempts on the problem?
– Clayton
Jul 19 at 16:03

Have you made any attempts on the problem?
– Clayton
Jul 19 at 16:03

Try l'Hopital's rule
– Green.H
Jul 19 at 16:03

Try l'Hopital's rule
– Green.H
Jul 19 at 16:03

2

2

Yes, l'Hopital doesn't help
– Dai Jinaid
Jul 19 at 16:10

Yes, l'Hopital doesn't help
– Dai Jinaid
Jul 19 at 16:10

add a comment | 

4 Answers
4

active

oldest

votes

up vote
1
down vote

By L'Hospital, twice,

$$frac(1+epsilon)^r+(1-epsilon)^r-2epsilon^2rtofrac(1+epsilon)^r-1-(1-epsilon)^r-12epsilon^2r-1tofracr-12r-1frac(1+epsilon)^r-2+(1-epsilon)^r-22epsilon^2r-2
\tofracr-12r-1epsilon^2-2rto0.$$

share|cite|improve this answer

edited Jul 19 at 16:32

answered Jul 19 at 16:27

Yves Daoust

111k665204

add a comment | 

up vote
1
down vote

The numerator is a smooth, even function so that by Taylor it is

$$aepsilon^2+o(epsilon^2)$$

which decays faster than $epsilon^2r$.

(More precisely $a=r(r-1)$, but this is unimportant.)

share|cite|improve this answer

edited Jul 19 at 16:49

answered Jul 19 at 16:40

Yves Daoust

111k665204

add a comment | 

up vote
0
down vote

With the definition of derivative:
beginalign
lim_epsilonrightarrow 0frac(1+epsilon)^r+(1-epsilon)^r-2epsilon^2r
&=left(lim_epsilonrightarrow 0frac(1+epsilon)^r-1epsilon + lim_epsilonrightarrow 0frac(1-epsilon)^r-1epsilonright)lim_epsilonrightarrow 0epsilon^1-2r\
&=left(((1+epsilon)^r)'Big|_epsilon=0 + ((1-epsilon)^r)'Big|_epsilon=0right)lim_epsilonrightarrow 0epsilon^1-2r\
&=2rtimes0\
&=0
endalign
only valid for $r<dfrac12$.

share|cite|improve this answer

edited Jul 19 at 16:20

answered Jul 19 at 16:16

Nosrati

19.6k41544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Shouldn't it be 1-2r instead of 2-2r?
  – Dai Jinaid
  Jul 19 at 16:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Unfortunately yes!
  – Nosrati
  Jul 19 at 16:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Apparently it is true for all r in (0,1)
  – Dai Jinaid
  Jul 19 at 16:23
  

  
  
  
  

add a comment | 

up vote
0
down vote

Use series expansion at $x=0$
beginalign
(1+x)^r &= sum_k=0^infty x^kbinomrk\
(1+x)^r+(1-x)^r-2 & =sum_k=0^infty x^kbinomrk + sum_k=0^infty (-x)^kbinomrk -2 \
&=2sum_k=1^infty x^2kbinomr2k\
frac(1+x)^r+(1-x)^r-2x^2r&=2sum_k=1^infty x^2k-2rbinomr2k
endalign
Since $rin (0,1)$, $2rin (0,2)$ and $2k-2r >0$ $forall kgeq 1$ so
$$
lim_xto 0frac(1+x)^r+(1-x)^r-2x^2r = 0
$$

share|cite|improve this answer

answered Jul 19 at 16:23

Rafael Gonzalez Lopez

652112

add a comment | 

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

By L'Hospital, twice,

$$frac(1+epsilon)^r+(1-epsilon)^r-2epsilon^2rtofrac(1+epsilon)^r-1-(1-epsilon)^r-12epsilon^2r-1tofracr-12r-1frac(1+epsilon)^r-2+(1-epsilon)^r-22epsilon^2r-2
\tofracr-12r-1epsilon^2-2rto0.$$

share|cite|improve this answer

edited Jul 19 at 16:32

answered Jul 19 at 16:27

Yves Daoust

111k665204

add a comment | 

up vote
1
down vote

By L'Hospital, twice,

$$frac(1+epsilon)^r+(1-epsilon)^r-2epsilon^2rtofrac(1+epsilon)^r-1-(1-epsilon)^r-12epsilon^2r-1tofracr-12r-1frac(1+epsilon)^r-2+(1-epsilon)^r-22epsilon^2r-2
\tofracr-12r-1epsilon^2-2rto0.$$

share|cite|improve this answer

edited Jul 19 at 16:32

answered Jul 19 at 16:27

Yves Daoust

111k665204

add a comment | 

up vote
1
down vote

up vote
1
down vote

By L'Hospital, twice,

$$frac(1+epsilon)^r+(1-epsilon)^r-2epsilon^2rtofrac(1+epsilon)^r-1-(1-epsilon)^r-12epsilon^2r-1tofracr-12r-1frac(1+epsilon)^r-2+(1-epsilon)^r-22epsilon^2r-2
\tofracr-12r-1epsilon^2-2rto0.$$

share|cite|improve this answer

edited Jul 19 at 16:32

answered Jul 19 at 16:27

Yves Daoust

111k665204

By L'Hospital, twice,

$$frac(1+epsilon)^r+(1-epsilon)^r-2epsilon^2rtofrac(1+epsilon)^r-1-(1-epsilon)^r-12epsilon^2r-1tofracr-12r-1frac(1+epsilon)^r-2+(1-epsilon)^r-22epsilon^2r-2
\tofracr-12r-1epsilon^2-2rto0.$$

share|cite|improve this answer

edited Jul 19 at 16:32

answered Jul 19 at 16:27

Yves Daoust

111k665204

share|cite|improve this answer

share|cite|improve this answer

edited Jul 19 at 16:32

edited Jul 19 at 16:32

edited Jul 19 at 16:32

answered Jul 19 at 16:27

Yves Daoust

111k665204

answered Jul 19 at 16:27

Yves Daoust

111k665204

answered Jul 19 at 16:27

Yves Daoust

111k665204

111k665204

add a comment | 

add a comment | 

up vote
1
down vote

The numerator is a smooth, even function so that by Taylor it is

$$aepsilon^2+o(epsilon^2)$$

which decays faster than $epsilon^2r$.

(More precisely $a=r(r-1)$, but this is unimportant.)

share|cite|improve this answer

edited Jul 19 at 16:49

answered Jul 19 at 16:40

Yves Daoust

111k665204

add a comment | 

up vote
1
down vote

The numerator is a smooth, even function so that by Taylor it is

$$aepsilon^2+o(epsilon^2)$$

which decays faster than $epsilon^2r$.

(More precisely $a=r(r-1)$, but this is unimportant.)

share|cite|improve this answer

edited Jul 19 at 16:49

answered Jul 19 at 16:40

Yves Daoust

111k665204

add a comment | 

up vote
1
down vote

up vote
1
down vote

The numerator is a smooth, even function so that by Taylor it is

$$aepsilon^2+o(epsilon^2)$$

which decays faster than $epsilon^2r$.

(More precisely $a=r(r-1)$, but this is unimportant.)

share|cite|improve this answer

edited Jul 19 at 16:49

answered Jul 19 at 16:40

Yves Daoust

111k665204

The numerator is a smooth, even function so that by Taylor it is

$$aepsilon^2+o(epsilon^2)$$

which decays faster than $epsilon^2r$.

(More precisely $a=r(r-1)$, but this is unimportant.)

share|cite|improve this answer

edited Jul 19 at 16:49

answered Jul 19 at 16:40

Yves Daoust

111k665204

share|cite|improve this answer

share|cite|improve this answer

edited Jul 19 at 16:49

edited Jul 19 at 16:49

edited Jul 19 at 16:49

answered Jul 19 at 16:40

Yves Daoust

111k665204

answered Jul 19 at 16:40

Yves Daoust

111k665204

answered Jul 19 at 16:40

Yves Daoust

111k665204

111k665204

add a comment | 

add a comment | 

up vote
0
down vote

With the definition of derivative:
beginalign
lim_epsilonrightarrow 0frac(1+epsilon)^r+(1-epsilon)^r-2epsilon^2r
&=left(lim_epsilonrightarrow 0frac(1+epsilon)^r-1epsilon + lim_epsilonrightarrow 0frac(1-epsilon)^r-1epsilonright)lim_epsilonrightarrow 0epsilon^1-2r\
&=left(((1+epsilon)^r)'Big|_epsilon=0 + ((1-epsilon)^r)'Big|_epsilon=0right)lim_epsilonrightarrow 0epsilon^1-2r\
&=2rtimes0\
&=0
endalign
only valid for $r<dfrac12$.

share|cite|improve this answer

edited Jul 19 at 16:20

answered Jul 19 at 16:16

Nosrati

19.6k41544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Shouldn't it be 1-2r instead of 2-2r?
  – Dai Jinaid
  Jul 19 at 16:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Unfortunately yes!
  – Nosrati
  Jul 19 at 16:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Apparently it is true for all r in (0,1)
  – Dai Jinaid
  Jul 19 at 16:23
  

  
  
  
  

add a comment | 

up vote
0
down vote

With the definition of derivative:
beginalign
lim_epsilonrightarrow 0frac(1+epsilon)^r+(1-epsilon)^r-2epsilon^2r
&=left(lim_epsilonrightarrow 0frac(1+epsilon)^r-1epsilon + lim_epsilonrightarrow 0frac(1-epsilon)^r-1epsilonright)lim_epsilonrightarrow 0epsilon^1-2r\
&=left(((1+epsilon)^r)'Big|_epsilon=0 + ((1-epsilon)^r)'Big|_epsilon=0right)lim_epsilonrightarrow 0epsilon^1-2r\
&=2rtimes0\
&=0
endalign
only valid for $r<dfrac12$.

share|cite|improve this answer

edited Jul 19 at 16:20

answered Jul 19 at 16:16

Nosrati

19.6k41544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Shouldn't it be 1-2r instead of 2-2r?
  – Dai Jinaid
  Jul 19 at 16:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Unfortunately yes!
  – Nosrati
  Jul 19 at 16:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Apparently it is true for all r in (0,1)
  – Dai Jinaid
  Jul 19 at 16:23
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

With the definition of derivative:
beginalign
lim_epsilonrightarrow 0frac(1+epsilon)^r+(1-epsilon)^r-2epsilon^2r
&=left(lim_epsilonrightarrow 0frac(1+epsilon)^r-1epsilon + lim_epsilonrightarrow 0frac(1-epsilon)^r-1epsilonright)lim_epsilonrightarrow 0epsilon^1-2r\
&=left(((1+epsilon)^r)'Big|_epsilon=0 + ((1-epsilon)^r)'Big|_epsilon=0right)lim_epsilonrightarrow 0epsilon^1-2r\
&=2rtimes0\
&=0
endalign
only valid for $r<dfrac12$.

share|cite|improve this answer

edited Jul 19 at 16:20

answered Jul 19 at 16:16

Nosrati

19.6k41544

With the definition of derivative:
beginalign
lim_epsilonrightarrow 0frac(1+epsilon)^r+(1-epsilon)^r-2epsilon^2r
&=left(lim_epsilonrightarrow 0frac(1+epsilon)^r-1epsilon + lim_epsilonrightarrow 0frac(1-epsilon)^r-1epsilonright)lim_epsilonrightarrow 0epsilon^1-2r\
&=left(((1+epsilon)^r)'Big|_epsilon=0 + ((1-epsilon)^r)'Big|_epsilon=0right)lim_epsilonrightarrow 0epsilon^1-2r\
&=2rtimes0\
&=0
endalign
only valid for $r<dfrac12$.

share|cite|improve this answer

edited Jul 19 at 16:20

answered Jul 19 at 16:16

Nosrati

19.6k41544

share|cite|improve this answer

share|cite|improve this answer

edited Jul 19 at 16:20

edited Jul 19 at 16:20

edited Jul 19 at 16:20

answered Jul 19 at 16:16

Nosrati

19.6k41544

answered Jul 19 at 16:16

Nosrati

19.6k41544

answered Jul 19 at 16:16

Nosrati

19.6k41544

19.6k41544

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Shouldn't it be 1-2r instead of 2-2r?
  – Dai Jinaid
  Jul 19 at 16:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Unfortunately yes!
  – Nosrati
  Jul 19 at 16:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Apparently it is true for all r in (0,1)
  – Dai Jinaid
  Jul 19 at 16:23
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Shouldn't it be 1-2r instead of 2-2r?
  – Dai Jinaid
  Jul 19 at 16:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Unfortunately yes!
  – Nosrati
  Jul 19 at 16:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Apparently it is true for all r in (0,1)
  – Dai Jinaid
  Jul 19 at 16:23
  

  
  
  
  

Shouldn't it be 1-2r instead of 2-2r?
– Dai Jinaid
Jul 19 at 16:20

Shouldn't it be 1-2r instead of 2-2r?
– Dai Jinaid
Jul 19 at 16:20

Unfortunately yes!
– Nosrati
Jul 19 at 16:23

Unfortunately yes!
– Nosrati
Jul 19 at 16:23

Apparently it is true for all r in (0,1)
– Dai Jinaid
Jul 19 at 16:23

Apparently it is true for all r in (0,1)
– Dai Jinaid
Jul 19 at 16:23

add a comment | 

up vote
0
down vote

Use series expansion at $x=0$
beginalign
(1+x)^r &= sum_k=0^infty x^kbinomrk\
(1+x)^r+(1-x)^r-2 & =sum_k=0^infty x^kbinomrk + sum_k=0^infty (-x)^kbinomrk -2 \
&=2sum_k=1^infty x^2kbinomr2k\
frac(1+x)^r+(1-x)^r-2x^2r&=2sum_k=1^infty x^2k-2rbinomr2k
endalign
Since $rin (0,1)$, $2rin (0,2)$ and $2k-2r >0$ $forall kgeq 1$ so
$$
lim_xto 0frac(1+x)^r+(1-x)^r-2x^2r = 0
$$

share|cite|improve this answer

answered Jul 19 at 16:23

Rafael Gonzalez Lopez

652112

add a comment | 

up vote
0
down vote

Use series expansion at $x=0$
beginalign
(1+x)^r &= sum_k=0^infty x^kbinomrk\
(1+x)^r+(1-x)^r-2 & =sum_k=0^infty x^kbinomrk + sum_k=0^infty (-x)^kbinomrk -2 \
&=2sum_k=1^infty x^2kbinomr2k\
frac(1+x)^r+(1-x)^r-2x^2r&=2sum_k=1^infty x^2k-2rbinomr2k
endalign
Since $rin (0,1)$, $2rin (0,2)$ and $2k-2r >0$ $forall kgeq 1$ so
$$
lim_xto 0frac(1+x)^r+(1-x)^r-2x^2r = 0
$$

share|cite|improve this answer

answered Jul 19 at 16:23

Rafael Gonzalez Lopez

652112

add a comment | 

up vote
0
down vote

up vote
0
down vote

Use series expansion at $x=0$
beginalign
(1+x)^r &= sum_k=0^infty x^kbinomrk\
(1+x)^r+(1-x)^r-2 & =sum_k=0^infty x^kbinomrk + sum_k=0^infty (-x)^kbinomrk -2 \
&=2sum_k=1^infty x^2kbinomr2k\
frac(1+x)^r+(1-x)^r-2x^2r&=2sum_k=1^infty x^2k-2rbinomr2k
endalign
Since $rin (0,1)$, $2rin (0,2)$ and $2k-2r >0$ $forall kgeq 1$ so
$$
lim_xto 0frac(1+x)^r+(1-x)^r-2x^2r = 0
$$

share|cite|improve this answer

answered Jul 19 at 16:23

Rafael Gonzalez Lopez

652112

Use series expansion at $x=0$
beginalign
(1+x)^r &= sum_k=0^infty x^kbinomrk\
(1+x)^r+(1-x)^r-2 & =sum_k=0^infty x^kbinomrk + sum_k=0^infty (-x)^kbinomrk -2 \
&=2sum_k=1^infty x^2kbinomr2k\
frac(1+x)^r+(1-x)^r-2x^2r&=2sum_k=1^infty x^2k-2rbinomr2k
endalign
Since $rin (0,1)$, $2rin (0,2)$ and $2k-2r >0$ $forall kgeq 1$ so
$$
lim_xto 0frac(1+x)^r+(1-x)^r-2x^2r = 0
$$

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answered Jul 19 at 16:23

Rafael Gonzalez Lopez

652112

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share|cite|improve this answer

answered Jul 19 at 16:23

Rafael Gonzalez Lopez

652112

answered Jul 19 at 16:23

Rafael Gonzalez Lopez

652112

answered Jul 19 at 16:23

Rafael Gonzalez Lopez

652112

652112

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