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Determine the value, when defined

0

$begingroup$

Determine the value, when defined, of $$lim_xtoinfty frac1x^2-lfloor xrfloor$$ where $lfloor xrfloor$ is the greatest integer.

I know $lim_xtoinfty lfloor xrfloor = infty$ and $lim_xtoinfty x^2$=$infty$ but how do I combine them in order to get $$lim_xtoinfty frac1x^2-[x]$$

limits analysis

share|cite|improve this question

edited Mar 27 at 16:13

user597188

asked Mar 27 at 16:00

user597188user597188

236

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What do you mean by $[x]$?
  $endgroup$
  – copper.hat
  Mar 27 at 16:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $lim_xtoinfty lfloor xrfloor =infty,$ not $-infty.$
  $endgroup$
  – Thomas Andrews
  Mar 27 at 16:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  copper.hat greatest integer
  $endgroup$
  – user597188
  Mar 27 at 16:12
  

  
  
  
  

add a comment | 

0

$begingroup$

Determine the value, when defined, of $$lim_xtoinfty frac1x^2-lfloor xrfloor$$ where $lfloor xrfloor$ is the greatest integer.

I know $lim_xtoinfty lfloor xrfloor = infty$ and $lim_xtoinfty x^2$=$infty$ but how do I combine them in order to get $$lim_xtoinfty frac1x^2-[x]$$

limits analysis

share|cite|improve this question

edited Mar 27 at 16:13

user597188

asked Mar 27 at 16:00

user597188user597188

236

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What do you mean by $[x]$?
  $endgroup$
  – copper.hat
  Mar 27 at 16:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $lim_xtoinfty lfloor xrfloor =infty,$ not $-infty.$
  $endgroup$
  – Thomas Andrews
  Mar 27 at 16:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  copper.hat greatest integer
  $endgroup$
  – user597188
  Mar 27 at 16:12
  

  
  
  
  

add a comment | 

$begingroup$

Determine the value, when defined, of $$lim_xtoinfty frac1x^2-lfloor xrfloor$$ where $lfloor xrfloor$ is the greatest integer.

I know $lim_xtoinfty lfloor xrfloor = infty$ and $lim_xtoinfty x^2$=$infty$ but how do I combine them in order to get $$lim_xtoinfty frac1x^2-[x]$$

limits analysis

share|cite|improve this question

edited Mar 27 at 16:13

user597188

asked Mar 27 at 16:00

user597188user597188

236

$endgroup$

share|cite|improve this question

edited Mar 27 at 16:13

user597188

asked Mar 27 at 16:00

user597188user597188

236

share|cite|improve this question

edited Mar 27 at 16:13

user597188

asked Mar 27 at 16:00

user597188user597188

236

asked Mar 27 at 16:00

user597188user597188

236

asked Mar 27 at 16:00

user597188user597188

236

3 Answers
3

active

oldest

votes

3

$begingroup$

Notice that $ x-1 leq [x]leq x$ so:

$$lim_xto inftyfrac1x^2-x+1leqlim_xto infty frac1x^2-[x]leq lim_xto infty frac1x^2-x$$

So:

$$0leqlim_xto infty frac1x^2-[x]leq 0$$

And:
$$lim_xto infty frac1x^2-[x]=0$$

:)

share|cite|improve this answer

answered Mar 27 at 16:12

EurekaEureka

907115

$endgroup$

add a comment | 

1

$begingroup$

We always have $lfloor xrfloorle x$.
For $x ge 2$ we have $x^2 over 2 ge x$.

Hence, if $x ge 2$, we have $x^2-lfloor xrfloor ge x^2 -x ge x^2 over 2$ and
so
$1 over x^2-lfloor xrfloor le 2 over x^2$.

share|cite|improve this answer

edited Mar 27 at 16:32

answered Mar 27 at 16:18

copper.hatcopper.hat

128k561161

$endgroup$

add a comment | 

0

$begingroup$

Is in not obvious that $x^2$ grows proportionally faster than $[x]$ so $x^2 - [x]to infty$?

To verify:

$x ge [x] > 0$ so $x^2 ge [x]^2$ and $frac 1 x^2 - [x] le frac 1[x]^2 - [x] = frac 1[x]([x] - 1)< frac 1([x]-1)^2to 0$.

To formalize:

If $epsilon > 0$ and we want $[x]- 1 > sqrtfrac 1epsilon$, which we can get if we choose $x > lceil sqrtfrac 1epsilon rceil + 1=N$.

Now if $x > N =lceil sqrtfrac 1epsilon rceil + 1$ then $[x] ge lceil sqrtfrac 1epsilon rceil + 1 ge sqrtfrac 1epsilon + 1$

$[x]- 1 ge sqrtfrac 1 epsilon$

$x^2 - [x]>[x]^2 - [x]=[x]([x]- 1)>([x]-1)^2 ge frac 1epsilon$

$frac 1x^2 -[x] < epsilon$.

So $limlimits_xto infty frac 1x^2 -[x] = infty$.

share|cite|improve this answer

answered Mar 27 at 16:38

fleabloodfleablood

1

$endgroup$

add a comment | 

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3

$begingroup$

Notice that $ x-1 leq [x]leq x$ so:

$$lim_xto inftyfrac1x^2-x+1leqlim_xto infty frac1x^2-[x]leq lim_xto infty frac1x^2-x$$

So:

$$0leqlim_xto infty frac1x^2-[x]leq 0$$

And:
$$lim_xto infty frac1x^2-[x]=0$$

:)

share|cite|improve this answer

answered Mar 27 at 16:12

EurekaEureka

907115

$endgroup$

add a comment | 

3

$begingroup$

Notice that $ x-1 leq [x]leq x$ so:

$$lim_xto inftyfrac1x^2-x+1leqlim_xto infty frac1x^2-[x]leq lim_xto infty frac1x^2-x$$

So:

$$0leqlim_xto infty frac1x^2-[x]leq 0$$

And:
$$lim_xto infty frac1x^2-[x]=0$$

:)

share|cite|improve this answer

answered Mar 27 at 16:12

EurekaEureka

907115

$endgroup$

add a comment | 

$begingroup$

Notice that $ x-1 leq [x]leq x$ so:

$$lim_xto inftyfrac1x^2-x+1leqlim_xto infty frac1x^2-[x]leq lim_xto infty frac1x^2-x$$

So:

$$0leqlim_xto infty frac1x^2-[x]leq 0$$

And:
$$lim_xto infty frac1x^2-[x]=0$$

:)

share|cite|improve this answer

answered Mar 27 at 16:12

EurekaEureka

907115

$endgroup$

share|cite|improve this answer

answered Mar 27 at 16:12

EurekaEureka

907115

answered Mar 27 at 16:12

EurekaEureka

907115

answered Mar 27 at 16:12

EurekaEureka

907115

1

$begingroup$

We always have $lfloor xrfloorle x$.
For $x ge 2$ we have $x^2 over 2 ge x$.

Hence, if $x ge 2$, we have $x^2-lfloor xrfloor ge x^2 -x ge x^2 over 2$ and
so
$1 over x^2-lfloor xrfloor le 2 over x^2$.

share|cite|improve this answer

edited Mar 27 at 16:32

answered Mar 27 at 16:18

copper.hatcopper.hat

128k561161

$endgroup$

add a comment | 

1

$begingroup$

We always have $lfloor xrfloorle x$.
For $x ge 2$ we have $x^2 over 2 ge x$.

Hence, if $x ge 2$, we have $x^2-lfloor xrfloor ge x^2 -x ge x^2 over 2$ and
so
$1 over x^2-lfloor xrfloor le 2 over x^2$.

share|cite|improve this answer

edited Mar 27 at 16:32

answered Mar 27 at 16:18

copper.hatcopper.hat

128k561161

$endgroup$

add a comment | 

$begingroup$

We always have $lfloor xrfloorle x$.
For $x ge 2$ we have $x^2 over 2 ge x$.

Hence, if $x ge 2$, we have $x^2-lfloor xrfloor ge x^2 -x ge x^2 over 2$ and
so
$1 over x^2-lfloor xrfloor le 2 over x^2$.

share|cite|improve this answer

edited Mar 27 at 16:32

answered Mar 27 at 16:18

copper.hatcopper.hat

128k561161

$endgroup$

share|cite|improve this answer

edited Mar 27 at 16:32

answered Mar 27 at 16:18

copper.hatcopper.hat

128k561161

answered Mar 27 at 16:18

copper.hatcopper.hat

128k561161

answered Mar 27 at 16:18

copper.hatcopper.hat

128k561161

0

$begingroup$

Is in not obvious that $x^2$ grows proportionally faster than $[x]$ so $x^2 - [x]to infty$?

To verify:

$x ge [x] > 0$ so $x^2 ge [x]^2$ and $frac 1 x^2 - [x] le frac 1[x]^2 - [x] = frac 1[x]([x] - 1)< frac 1([x]-1)^2to 0$.

To formalize:

If $epsilon > 0$ and we want $[x]- 1 > sqrtfrac 1epsilon$, which we can get if we choose $x > lceil sqrtfrac 1epsilon rceil + 1=N$.

Now if $x > N =lceil sqrtfrac 1epsilon rceil + 1$ then $[x] ge lceil sqrtfrac 1epsilon rceil + 1 ge sqrtfrac 1epsilon + 1$

$[x]- 1 ge sqrtfrac 1 epsilon$

$x^2 - [x]>[x]^2 - [x]=[x]([x]- 1)>([x]-1)^2 ge frac 1epsilon$

$frac 1x^2 -[x] < epsilon$.

So $limlimits_xto infty frac 1x^2 -[x] = infty$.

share|cite|improve this answer

answered Mar 27 at 16:38

fleabloodfleablood

1

$endgroup$

add a comment | 

0

$begingroup$

Is in not obvious that $x^2$ grows proportionally faster than $[x]$ so $x^2 - [x]to infty$?

To verify:

$x ge [x] > 0$ so $x^2 ge [x]^2$ and $frac 1 x^2 - [x] le frac 1[x]^2 - [x] = frac 1[x]([x] - 1)< frac 1([x]-1)^2to 0$.

To formalize:

If $epsilon > 0$ and we want $[x]- 1 > sqrtfrac 1epsilon$, which we can get if we choose $x > lceil sqrtfrac 1epsilon rceil + 1=N$.

Now if $x > N =lceil sqrtfrac 1epsilon rceil + 1$ then $[x] ge lceil sqrtfrac 1epsilon rceil + 1 ge sqrtfrac 1epsilon + 1$

$[x]- 1 ge sqrtfrac 1 epsilon$

$x^2 - [x]>[x]^2 - [x]=[x]([x]- 1)>([x]-1)^2 ge frac 1epsilon$

$frac 1x^2 -[x] < epsilon$.

So $limlimits_xto infty frac 1x^2 -[x] = infty$.

share|cite|improve this answer

answered Mar 27 at 16:38

fleabloodfleablood

1

$endgroup$

add a comment | 

$begingroup$

Is in not obvious that $x^2$ grows proportionally faster than $[x]$ so $x^2 - [x]to infty$?

To verify:

$x ge [x] > 0$ so $x^2 ge [x]^2$ and $frac 1 x^2 - [x] le frac 1[x]^2 - [x] = frac 1[x]([x] - 1)< frac 1([x]-1)^2to 0$.

To formalize:

If $epsilon > 0$ and we want $[x]- 1 > sqrtfrac 1epsilon$, which we can get if we choose $x > lceil sqrtfrac 1epsilon rceil + 1=N$.

Now if $x > N =lceil sqrtfrac 1epsilon rceil + 1$ then $[x] ge lceil sqrtfrac 1epsilon rceil + 1 ge sqrtfrac 1epsilon + 1$

$[x]- 1 ge sqrtfrac 1 epsilon$

$x^2 - [x]>[x]^2 - [x]=[x]([x]- 1)>([x]-1)^2 ge frac 1epsilon$

$frac 1x^2 -[x] < epsilon$.

So $limlimits_xto infty frac 1x^2 -[x] = infty$.

share|cite|improve this answer

answered Mar 27 at 16:38

fleabloodfleablood

1

$endgroup$

share|cite|improve this answer

answered Mar 27 at 16:38

fleabloodfleablood

1

answered Mar 27 at 16:38

fleabloodfleablood

1

answered Mar 27 at 16:38

fleabloodfleablood

1