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Calculate $lim_{x rightarrow 0} frac{e^{x}-x-1}{1-cos x}$

1

$begingroup$

I know how to do this task and my answer is $1$.

However I think that my way is a long way to the goal and I want to try do this task using Taylor's theorem with the rest in the form of Peano.

Unfortunately I do not fully understand how I can do it.

Can you help me?

real-analysis

share|cite|improve this question

edited Mar 26 at 10:10

Ernie060

2,950719

asked Mar 26 at 10:04

MP3129MP3129

900211

$endgroup$

add a comment | 

1

$begingroup$

I know how to do this task and my answer is $1$.

However I think that my way is a long way to the goal and I want to try do this task using Taylor's theorem with the rest in the form of Peano.

Unfortunately I do not fully understand how I can do it.

Can you help me?

real-analysis

share|cite|improve this question

edited Mar 26 at 10:10

Ernie060

2,950719

asked Mar 26 at 10:04

MP3129MP3129

900211

$endgroup$

add a comment | 

$begingroup$

I know how to do this task and my answer is $1$.

However I think that my way is a long way to the goal and I want to try do this task using Taylor's theorem with the rest in the form of Peano.

Unfortunately I do not fully understand how I can do it.

Can you help me?

real-analysis

share|cite|improve this question

edited Mar 26 at 10:10

Ernie060

2,950719

asked Mar 26 at 10:04

MP3129MP3129

900211

$endgroup$

share|cite|improve this question

edited Mar 26 at 10:10

Ernie060

2,950719

asked Mar 26 at 10:04

MP3129MP3129

900211

share|cite|improve this question

edited Mar 26 at 10:10

Ernie060

2,950719

asked Mar 26 at 10:04

MP3129MP3129

900211

edited Mar 26 at 10:10

Ernie060

2,950719

edited Mar 26 at 10:10

Ernie060

2,950719

asked Mar 26 at 10:04

MP3129MP3129

900211

asked Mar 26 at 10:04

MP3129MP3129

900211

5 Answers
5

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oldest

votes

3

$begingroup$

Using

• 
  $e^x = 1+x+frac{x^2}{2} + o(x^2)$ and


• $cos x = 1 - frac{x^2}{2} + o(x^2)$

you get
$$frac{e^{x}-x-1}{1-cos x} = frac{frac{x^2}{2} + o(x^2)}{frac{x^2}{2} + o(x^2)}stackrel{x to 0}{longrightarrow}1$$

share|cite|improve this answer

answered Mar 26 at 10:10

trancelocationtrancelocation

14.5k1929

$endgroup$

add a comment | 

3

$begingroup$

Using expansions up to order $2$:
$$lim_{xto 0} frac{e^x-x-1}{1-cos x} = lim_{x to 0}frac{1+x+x^2/2+o(x^2)-x-1}{1-1+x^2/2+o(x^2)} = lim_{x to 0} frac{1/2+o(x^2)/x^2}{1/2+o(x^2)/x^2} = 1. $$

share|cite|improve this answer

answered Mar 26 at 10:10

GibbsGibbs

5,4323927

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add a comment | 

1

$begingroup$

The simpest way is to apply L'Hopital's Rule twice.

share|cite|improve this answer

answered Mar 26 at 10:06

Kavi Rama MurthyKavi Rama Murthy

76.6k53471

$endgroup$

add a comment | 

1

$begingroup$

Using l'Hospital: $$lim_{xrightarrow 0}frac{exp(x)-x-1}{1-cos(x)}=lim_{xrightarrow 0}frac{exp(x)-1}{sin(x)}=lim_{xrightarrow 0}frac{exp(x)}{cos(x)}=frac{exp(0)}{cos(0)}=1.$$

share|cite|improve this answer

answered Mar 26 at 10:08

st.mathst.math

1,278115

$endgroup$

add a comment | 

1

$begingroup$

$$lim_{xto0}dfrac{e^x-1-x}{1-cos x}=lim_{xto0}dfrac{e^x-1-x}{x^2}lim_{xto0}left(dfrac x{sin x}right)^2lim_{xto0}(1+cos x)$$

For the first limit use Are all limits solvable without L'Hôpital Rule or Series Expansion

share|cite|improve this answer

answered Mar 26 at 10:11

lab bhattacharjeelab bhattacharjee

229k15159279

$endgroup$

add a comment | 

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3

$begingroup$

Using

• 
  $e^x = 1+x+frac{x^2}{2} + o(x^2)$ and


• $cos x = 1 - frac{x^2}{2} + o(x^2)$

you get
$$frac{e^{x}-x-1}{1-cos x} = frac{frac{x^2}{2} + o(x^2)}{frac{x^2}{2} + o(x^2)}stackrel{x to 0}{longrightarrow}1$$

share|cite|improve this answer

answered Mar 26 at 10:10

trancelocationtrancelocation

14.5k1929

$endgroup$

add a comment | 

3

$begingroup$

Using

• 
  $e^x = 1+x+frac{x^2}{2} + o(x^2)$ and


• $cos x = 1 - frac{x^2}{2} + o(x^2)$

you get
$$frac{e^{x}-x-1}{1-cos x} = frac{frac{x^2}{2} + o(x^2)}{frac{x^2}{2} + o(x^2)}stackrel{x to 0}{longrightarrow}1$$

share|cite|improve this answer

answered Mar 26 at 10:10

trancelocationtrancelocation

14.5k1929

$endgroup$

add a comment | 

$begingroup$

Using

• 
  $e^x = 1+x+frac{x^2}{2} + o(x^2)$ and


• $cos x = 1 - frac{x^2}{2} + o(x^2)$

you get
$$frac{e^{x}-x-1}{1-cos x} = frac{frac{x^2}{2} + o(x^2)}{frac{x^2}{2} + o(x^2)}stackrel{x to 0}{longrightarrow}1$$

share|cite|improve this answer

answered Mar 26 at 10:10

trancelocationtrancelocation

14.5k1929

$endgroup$

share|cite|improve this answer

answered Mar 26 at 10:10

trancelocationtrancelocation

14.5k1929

answered Mar 26 at 10:10

trancelocationtrancelocation

14.5k1929

answered Mar 26 at 10:10

trancelocationtrancelocation

14.5k1929

3

$begingroup$

Using expansions up to order $2$:
$$lim_{xto 0} frac{e^x-x-1}{1-cos x} = lim_{x to 0}frac{1+x+x^2/2+o(x^2)-x-1}{1-1+x^2/2+o(x^2)} = lim_{x to 0} frac{1/2+o(x^2)/x^2}{1/2+o(x^2)/x^2} = 1. $$

share|cite|improve this answer

answered Mar 26 at 10:10

GibbsGibbs

5,4323927

$endgroup$

add a comment | 

3

$begingroup$

Using expansions up to order $2$:
$$lim_{xto 0} frac{e^x-x-1}{1-cos x} = lim_{x to 0}frac{1+x+x^2/2+o(x^2)-x-1}{1-1+x^2/2+o(x^2)} = lim_{x to 0} frac{1/2+o(x^2)/x^2}{1/2+o(x^2)/x^2} = 1. $$

share|cite|improve this answer

answered Mar 26 at 10:10

GibbsGibbs

5,4323927

$endgroup$

add a comment | 

$begingroup$

Using expansions up to order $2$:
$$lim_{xto 0} frac{e^x-x-1}{1-cos x} = lim_{x to 0}frac{1+x+x^2/2+o(x^2)-x-1}{1-1+x^2/2+o(x^2)} = lim_{x to 0} frac{1/2+o(x^2)/x^2}{1/2+o(x^2)/x^2} = 1. $$

share|cite|improve this answer

answered Mar 26 at 10:10

GibbsGibbs

5,4323927

$endgroup$

share|cite|improve this answer

answered Mar 26 at 10:10

GibbsGibbs

5,4323927

answered Mar 26 at 10:10

GibbsGibbs

5,4323927

answered Mar 26 at 10:10

GibbsGibbs

5,4323927

1

$begingroup$

The simpest way is to apply L'Hopital's Rule twice.

share|cite|improve this answer

answered Mar 26 at 10:06

Kavi Rama MurthyKavi Rama Murthy

76.6k53471

$endgroup$

add a comment | 

1

$begingroup$

The simpest way is to apply L'Hopital's Rule twice.

share|cite|improve this answer

answered Mar 26 at 10:06

Kavi Rama MurthyKavi Rama Murthy

76.6k53471

$endgroup$

add a comment | 

$begingroup$

The simpest way is to apply L'Hopital's Rule twice.

share|cite|improve this answer

answered Mar 26 at 10:06

Kavi Rama MurthyKavi Rama Murthy

76.6k53471

$endgroup$

share|cite|improve this answer

answered Mar 26 at 10:06

Kavi Rama MurthyKavi Rama Murthy

76.6k53471

answered Mar 26 at 10:06

Kavi Rama MurthyKavi Rama Murthy

76.6k53471

answered Mar 26 at 10:06

Kavi Rama MurthyKavi Rama Murthy

76.6k53471

1

$begingroup$

Using l'Hospital: $$lim_{xrightarrow 0}frac{exp(x)-x-1}{1-cos(x)}=lim_{xrightarrow 0}frac{exp(x)-1}{sin(x)}=lim_{xrightarrow 0}frac{exp(x)}{cos(x)}=frac{exp(0)}{cos(0)}=1.$$

share|cite|improve this answer

answered Mar 26 at 10:08

st.mathst.math

1,278115

$endgroup$

add a comment | 

1

$begingroup$

Using l'Hospital: $$lim_{xrightarrow 0}frac{exp(x)-x-1}{1-cos(x)}=lim_{xrightarrow 0}frac{exp(x)-1}{sin(x)}=lim_{xrightarrow 0}frac{exp(x)}{cos(x)}=frac{exp(0)}{cos(0)}=1.$$

share|cite|improve this answer

answered Mar 26 at 10:08

st.mathst.math

1,278115

$endgroup$

add a comment | 

$begingroup$

Using l'Hospital: $$lim_{xrightarrow 0}frac{exp(x)-x-1}{1-cos(x)}=lim_{xrightarrow 0}frac{exp(x)-1}{sin(x)}=lim_{xrightarrow 0}frac{exp(x)}{cos(x)}=frac{exp(0)}{cos(0)}=1.$$

share|cite|improve this answer

answered Mar 26 at 10:08

st.mathst.math

1,278115

$endgroup$

share|cite|improve this answer

answered Mar 26 at 10:08

st.mathst.math

1,278115

answered Mar 26 at 10:08

st.mathst.math

1,278115

answered Mar 26 at 10:08

st.mathst.math

1,278115

1

$begingroup$

$$lim_{xto0}dfrac{e^x-1-x}{1-cos x}=lim_{xto0}dfrac{e^x-1-x}{x^2}lim_{xto0}left(dfrac x{sin x}right)^2lim_{xto0}(1+cos x)$$

For the first limit use Are all limits solvable without L'Hôpital Rule or Series Expansion

share|cite|improve this answer

answered Mar 26 at 10:11

lab bhattacharjeelab bhattacharjee

229k15159279

$endgroup$

add a comment | 

1

$begingroup$

$$lim_{xto0}dfrac{e^x-1-x}{1-cos x}=lim_{xto0}dfrac{e^x-1-x}{x^2}lim_{xto0}left(dfrac x{sin x}right)^2lim_{xto0}(1+cos x)$$

For the first limit use Are all limits solvable without L'Hôpital Rule or Series Expansion

share|cite|improve this answer

answered Mar 26 at 10:11

lab bhattacharjeelab bhattacharjee

229k15159279

$endgroup$

add a comment | 

$begingroup$

$$lim_{xto0}dfrac{e^x-1-x}{1-cos x}=lim_{xto0}dfrac{e^x-1-x}{x^2}lim_{xto0}left(dfrac x{sin x}right)^2lim_{xto0}(1+cos x)$$

For the first limit use Are all limits solvable without L'Hôpital Rule or Series Expansion

share|cite|improve this answer

answered Mar 26 at 10:11

lab bhattacharjeelab bhattacharjee

229k15159279

$endgroup$

share|cite|improve this answer

answered Mar 26 at 10:11

lab bhattacharjeelab bhattacharjee

229k15159279

answered Mar 26 at 10:11

lab bhattacharjeelab bhattacharjee

229k15159279

answered Mar 26 at 10:11

lab bhattacharjeelab bhattacharjee

229k15159279