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Find $limlimits_{xtoinfty} frac{1}{x}int_{0}^{x} frac{1}{2+cos t}dt$.

Find the limit : $limlimits_{nrightarrowinfty}int_{n}^{n+7}frac{sin{x}}{x},mathrm dx$Could we solve $int_{0}^{infty}sin(x)dx$ and what does it say about $lim_{xtoinfty}cos(x)$?Find $ limlimits_{n rightarrow infty} int_{0}^{1} left(1+ frac{x}{n}right)^n dx$How to find: $ limlimits_{nrightarrowinfty} leftlfloor frac{-1}{n}rightrfloor $When is $int_{-infty}^infty f(x) dx = limlimits_{n to infty} int_{-n}^n f(x) dx$?$limlimits_{n rightarrow +infty} frac{ln(1+n+n^3)-3ln(n)}{n(1-cos(1/n^2))}$$limlimits_{n rightarrow +infty} frac{sumlimits_{k=1}^{n} sqrt[k] {k} }{n}= 1$Having the sequence $(a_{n})_{ngeq1}$, $a_{n}=int_{0}^{1} x^{n}(1-x)^{n}dx$, find $limlimits_{n{rightarrow}infty} frac{a_{n+1}}{a_{n}}$Find$ limlimits_{nrightarrowinfty}frac{x_{n}}{n}$Find the limit of $limlimits_{ntoinfty}frac{1}{n^2}sumlimits_{k=1}^{n}karctan{big(frac{pk-p+1}{pn}big)}$

2 Answers
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4

$begingroup$

The function
$$ f(t)=frac{1}{2+cos t} $$
is periodic with period $2pi$. Thus, we have that
$$int_0^x f(t),dt = N int_0^{2pi}f(t),dt + int_0^{x- 2pi N}f(t),dt $$
with
$$ N = leftlfloor frac{x}{2pi} rightrfloor,.$$
The second term is bounded by a constant,
$$left|int_0^{x- 2pi N}f(t),dt right| leq int_0^{2pi}|f(t)|,dt,.$$

Thus, we have that
$$lim_{xtoinfty} frac{1}{x} int_0^x f(t),dt = frac{1}{2pi}left(lim_{xtoinfty} frac{2pi}{x} leftlfloor frac{x}{2pi} rightrfloor right)int_0^{2pi}f(t),dt =frac{1}{2pi} int_0^{2pi}f(t),dt,.$$

The last integral can be easily evaluated and we obtain
$$ lim_{xtoinfty} frac{1}{x} int_0^x f(t),dt = frac{1}{2pi} int_0^{2pi}f(t),dt = frac{1}{sqrt{3}},.$$

share|cite|improve this answer

answered Mar 14 at 20:37

FabianFabian

20k3774

$endgroup$

add a comment | 

1

$begingroup$

As $-1le cos(t)le 1$ for any $t$,

$$frac{1}{2+cos(t)}ge frac{1}{3}$$

So

$$
int_0^x frac{1}{2+cos(t)} dt ge frac{x}{3}
$$

share|cite|improve this answer

answered Mar 14 at 20:14

ReveillarkReveillark

4,742822

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This may be useful, but does not provide me with an answer for the given limit.
  $endgroup$
  – user651754
  Mar 14 at 20:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user651754 Wouldn't the integral in question tend to $infty$?
  $endgroup$
  – Reveillark
  Mar 14 at 20:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I did a mistake by not adding everything given. Otherwise, the l'Hospital idea would have had no sense.
  $endgroup$
  – user651754
  Mar 14 at 20:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This only shows that the given limit is greater than or equal to $frac{1}{3}$.
  $endgroup$
  – aleden
  Mar 14 at 21:56
  

  
  
  
  

add a comment | 

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4

$begingroup$

The function
$$ f(t)=frac{1}{2+cos t} $$
is periodic with period $2pi$. Thus, we have that
$$int_0^x f(t),dt = N int_0^{2pi}f(t),dt + int_0^{x- 2pi N}f(t),dt $$
with
$$ N = leftlfloor frac{x}{2pi} rightrfloor,.$$
The second term is bounded by a constant,
$$left|int_0^{x- 2pi N}f(t),dt right| leq int_0^{2pi}|f(t)|,dt,.$$

Thus, we have that
$$lim_{xtoinfty} frac{1}{x} int_0^x f(t),dt = frac{1}{2pi}left(lim_{xtoinfty} frac{2pi}{x} leftlfloor frac{x}{2pi} rightrfloor right)int_0^{2pi}f(t),dt =frac{1}{2pi} int_0^{2pi}f(t),dt,.$$

The last integral can be easily evaluated and we obtain
$$ lim_{xtoinfty} frac{1}{x} int_0^x f(t),dt = frac{1}{2pi} int_0^{2pi}f(t),dt = frac{1}{sqrt{3}},.$$

share|cite|improve this answer

answered Mar 14 at 20:37

FabianFabian

20k3774

$endgroup$

add a comment | 

4

$begingroup$

The function
$$ f(t)=frac{1}{2+cos t} $$
is periodic with period $2pi$. Thus, we have that
$$int_0^x f(t),dt = N int_0^{2pi}f(t),dt + int_0^{x- 2pi N}f(t),dt $$
with
$$ N = leftlfloor frac{x}{2pi} rightrfloor,.$$
The second term is bounded by a constant,
$$left|int_0^{x- 2pi N}f(t),dt right| leq int_0^{2pi}|f(t)|,dt,.$$

Thus, we have that
$$lim_{xtoinfty} frac{1}{x} int_0^x f(t),dt = frac{1}{2pi}left(lim_{xtoinfty} frac{2pi}{x} leftlfloor frac{x}{2pi} rightrfloor right)int_0^{2pi}f(t),dt =frac{1}{2pi} int_0^{2pi}f(t),dt,.$$

The last integral can be easily evaluated and we obtain
$$ lim_{xtoinfty} frac{1}{x} int_0^x f(t),dt = frac{1}{2pi} int_0^{2pi}f(t),dt = frac{1}{sqrt{3}},.$$

share|cite|improve this answer

answered Mar 14 at 20:37

FabianFabian

20k3774

$endgroup$

add a comment | 

$begingroup$

The function
$$ f(t)=frac{1}{2+cos t} $$
is periodic with period $2pi$. Thus, we have that
$$int_0^x f(t),dt = N int_0^{2pi}f(t),dt + int_0^{x- 2pi N}f(t),dt $$
with
$$ N = leftlfloor frac{x}{2pi} rightrfloor,.$$
The second term is bounded by a constant,
$$left|int_0^{x- 2pi N}f(t),dt right| leq int_0^{2pi}|f(t)|,dt,.$$

Thus, we have that
$$lim_{xtoinfty} frac{1}{x} int_0^x f(t),dt = frac{1}{2pi}left(lim_{xtoinfty} frac{2pi}{x} leftlfloor frac{x}{2pi} rightrfloor right)int_0^{2pi}f(t),dt =frac{1}{2pi} int_0^{2pi}f(t),dt,.$$

The last integral can be easily evaluated and we obtain
$$ lim_{xtoinfty} frac{1}{x} int_0^x f(t),dt = frac{1}{2pi} int_0^{2pi}f(t),dt = frac{1}{sqrt{3}},.$$

share|cite|improve this answer

answered Mar 14 at 20:37

FabianFabian

20k3774

$endgroup$

share|cite|improve this answer

answered Mar 14 at 20:37

FabianFabian

20k3774

answered Mar 14 at 20:37

FabianFabian

20k3774

answered Mar 14 at 20:37

FabianFabian

20k3774

1

$begingroup$

As $-1le cos(t)le 1$ for any $t$,

$$frac{1}{2+cos(t)}ge frac{1}{3}$$

So

$$
int_0^x frac{1}{2+cos(t)} dt ge frac{x}{3}
$$

share|cite|improve this answer

answered Mar 14 at 20:14

ReveillarkReveillark

4,742822

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This may be useful, but does not provide me with an answer for the given limit.
  $endgroup$
  – user651754
  Mar 14 at 20:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user651754 Wouldn't the integral in question tend to $infty$?
  $endgroup$
  – Reveillark
  Mar 14 at 20:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I did a mistake by not adding everything given. Otherwise, the l'Hospital idea would have had no sense.
  $endgroup$
  – user651754
  Mar 14 at 20:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This only shows that the given limit is greater than or equal to $frac{1}{3}$.
  $endgroup$
  – aleden
  Mar 14 at 21:56
  

  
  
  
  

add a comment | 

1

$begingroup$

As $-1le cos(t)le 1$ for any $t$,

$$frac{1}{2+cos(t)}ge frac{1}{3}$$

So

$$
int_0^x frac{1}{2+cos(t)} dt ge frac{x}{3}
$$

share|cite|improve this answer

answered Mar 14 at 20:14

ReveillarkReveillark

4,742822

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This may be useful, but does not provide me with an answer for the given limit.
  $endgroup$
  – user651754
  Mar 14 at 20:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user651754 Wouldn't the integral in question tend to $infty$?
  $endgroup$
  – Reveillark
  Mar 14 at 20:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I did a mistake by not adding everything given. Otherwise, the l'Hospital idea would have had no sense.
  $endgroup$
  – user651754
  Mar 14 at 20:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This only shows that the given limit is greater than or equal to $frac{1}{3}$.
  $endgroup$
  – aleden
  Mar 14 at 21:56
  

  
  
  
  

add a comment | 

$begingroup$

As $-1le cos(t)le 1$ for any $t$,

$$frac{1}{2+cos(t)}ge frac{1}{3}$$

So

$$
int_0^x frac{1}{2+cos(t)} dt ge frac{x}{3}
$$

share|cite|improve this answer

answered Mar 14 at 20:14

ReveillarkReveillark

4,742822

$endgroup$

share|cite|improve this answer

answered Mar 14 at 20:14

ReveillarkReveillark

4,742822

answered Mar 14 at 20:14

ReveillarkReveillark

4,742822

answered Mar 14 at 20:14

ReveillarkReveillark

4,742822