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Find the limit or prove that it does not exist

Real Analysis Prove that Limit Does not Existevaluate the limit or show that it doesn't existDoes this iterated limit exist?Prove that the limit doesn't exist of the combination of sines and cosinesFinding. That a two variable limit does not exist troubleWhy does the limit $lim_x to 0 frac1-cos^3 xxsin 2x$ exist?Why does limit $lim_(x,y) to (0,0)frac2xx^2+x+y^2$ not exist?Determining if a limit does not exist.Prove the limit does not exist in multidimensionCompute each limit or state that it does not exist

3

1

$begingroup$

Find the limit or prove that it does not exist
$$lim_(x, space y) to (0, space 0) f(x, y) $$ where $$f(x, y) = fracx^5-y^5x^4-2x^2y^2+y^4$$

---

Iterated limit $displaystylelim_y to 0 space displaystylelim_x to 0 space f(x, y) = displaystylelim_x to 0 space displaystylelim_y to 0 space f(x, y) = 0$, but it doesn't mean that $displaystylelim_(x, space y) to (0, space 0) f(x, y) = 0 $.

I've also tried substitution $x = r cdot cos phi, space y = r cdot sin phi$, which gave me $ fracr(cos^5phi - sin^5phi)(cos^2phi-sin^2phi)^2 $. If $x to 0$ and $y to 0$, then $r to 0$. So, $displaystylelim_r to 0 space fracr(cos^5phi - sin^5phi)(cos^2phi-sin^2phi)^2 = 0$.

But I have a feeling that original limit doesn't exist and WolframAlpha says that too and I'm stuck here. If my assumption is correct, how to prove it properly?

calculus limits multivariable-calculus

share|cite|improve this question

asked 2 hours ago

Nikita GubanovNikita Gubanov

715

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Check the plot of the function to get an idea of curves y(x) along which the limit should be different, and then compute said limit along those two curves.For instance, if you choose $phi=pi/4$, you get 0/0, so the limit may not be $0$.
  $endgroup$
  – xihiro
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This function is not defined in any neighbourhood of $(0,0)$ (minus the origin).
  $endgroup$
  – Bernard
  2 hours ago
  

  
  
  
  

add a comment | 

3

1

$begingroup$

Find the limit or prove that it does not exist
$$lim_(x, space y) to (0, space 0) f(x, y) $$ where $$f(x, y) = fracx^5-y^5x^4-2x^2y^2+y^4$$

---

Iterated limit $displaystylelim_y to 0 space displaystylelim_x to 0 space f(x, y) = displaystylelim_x to 0 space displaystylelim_y to 0 space f(x, y) = 0$, but it doesn't mean that $displaystylelim_(x, space y) to (0, space 0) f(x, y) = 0 $.

I've also tried substitution $x = r cdot cos phi, space y = r cdot sin phi$, which gave me $ fracr(cos^5phi - sin^5phi)(cos^2phi-sin^2phi)^2 $. If $x to 0$ and $y to 0$, then $r to 0$. So, $displaystylelim_r to 0 space fracr(cos^5phi - sin^5phi)(cos^2phi-sin^2phi)^2 = 0$.

But I have a feeling that original limit doesn't exist and WolframAlpha says that too and I'm stuck here. If my assumption is correct, how to prove it properly?

calculus limits multivariable-calculus

share|cite|improve this question

asked 2 hours ago

Nikita GubanovNikita Gubanov

715

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Check the plot of the function to get an idea of curves y(x) along which the limit should be different, and then compute said limit along those two curves.For instance, if you choose $phi=pi/4$, you get 0/0, so the limit may not be $0$.
  $endgroup$
  – xihiro
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This function is not defined in any neighbourhood of $(0,0)$ (minus the origin).
  $endgroup$
  – Bernard
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Find the limit or prove that it does not exist
$$lim_(x, space y) to (0, space 0) f(x, y) $$ where $$f(x, y) = fracx^5-y^5x^4-2x^2y^2+y^4$$

---

Iterated limit $displaystylelim_y to 0 space displaystylelim_x to 0 space f(x, y) = displaystylelim_x to 0 space displaystylelim_y to 0 space f(x, y) = 0$, but it doesn't mean that $displaystylelim_(x, space y) to (0, space 0) f(x, y) = 0 $.

I've also tried substitution $x = r cdot cos phi, space y = r cdot sin phi$, which gave me $ fracr(cos^5phi - sin^5phi)(cos^2phi-sin^2phi)^2 $. If $x to 0$ and $y to 0$, then $r to 0$. So, $displaystylelim_r to 0 space fracr(cos^5phi - sin^5phi)(cos^2phi-sin^2phi)^2 = 0$.

But I have a feeling that original limit doesn't exist and WolframAlpha says that too and I'm stuck here. If my assumption is correct, how to prove it properly?

calculus limits multivariable-calculus

share|cite|improve this question

asked 2 hours ago

Nikita GubanovNikita Gubanov

715

$endgroup$

share|cite|improve this question

asked 2 hours ago

Nikita GubanovNikita Gubanov

715

share|cite|improve this question

asked 2 hours ago

Nikita GubanovNikita Gubanov

715

asked 2 hours ago

Nikita GubanovNikita Gubanov

715

asked 2 hours ago

Nikita GubanovNikita Gubanov

715

3 Answers
3

active

oldest

votes

3

$begingroup$

The limit doesn't exist. If $ninmathbb N$, then$$fleft(sqrtfrac1n^2+frac1n^4,frac1nright)=left(left(frac1n^2+frac1n^4right)^5/2-frac1n^5right)n^8$$and$$lim_ntoinftyleft(left(frac1n^2+frac1n^4right)^5/2-frac1n^5right)n^8=infty.$$

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

José Carlos SantosJosé Carlos Santos

180k24140254

$endgroup$

add a comment | 

2

$begingroup$

I think your substitution gives you a hint. The angles $tanphi =pm 1$ are the problematic ones, i.e., the limit along the lines $pm x=y$. You can check that along these lines there's a divergence.

share|cite|improve this answer

edited 39 mins ago

answered 1 hour ago

olaphusolaphus

763

$endgroup$

add a comment | 

0

$begingroup$

In the posting, $$ f(r,phi)=fracr(cos^5phi -sin^5phi
)(cos^2phi -sin^2phi)^2 = underbracefracrcos
phi-sin phi_=gunderbracefracsum_i=1^4
sin^iphicos^4-iphi (cos phi+sin phi)^2 _=h$$

When $f(r,0)=r$, so $lim_r f=0$.

Further, $lim_phi rightarrow
pi/4 h $ exists. We have a sequence $phi_n<pi/4,
phi_nrightarrow pi/4$ s.t. $ cos phi_n > sin phi_n$. We let
$$r_n =sqrtcos phi_n - sin phi_n.$$ Hence $$ g(r_n,phi_n) =
frac1sqrt cos phi_n - sin phi_n rightarrow infty$$
so that $f(r_n,phi_n)$ goes to $infty$.

share|cite|improve this answer

answered 22 mins ago

HK LeeHK Lee

14.2k52363

$endgroup$

add a comment | 

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3

$begingroup$

The limit doesn't exist. If $ninmathbb N$, then$$fleft(sqrtfrac1n^2+frac1n^4,frac1nright)=left(left(frac1n^2+frac1n^4right)^5/2-frac1n^5right)n^8$$and$$lim_ntoinftyleft(left(frac1n^2+frac1n^4right)^5/2-frac1n^5right)n^8=infty.$$

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

José Carlos SantosJosé Carlos Santos

180k24140254

$endgroup$

add a comment | 

3

$begingroup$

The limit doesn't exist. If $ninmathbb N$, then$$fleft(sqrtfrac1n^2+frac1n^4,frac1nright)=left(left(frac1n^2+frac1n^4right)^5/2-frac1n^5right)n^8$$and$$lim_ntoinftyleft(left(frac1n^2+frac1n^4right)^5/2-frac1n^5right)n^8=infty.$$

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

José Carlos SantosJosé Carlos Santos

180k24140254

$endgroup$

add a comment | 

$begingroup$

The limit doesn't exist. If $ninmathbb N$, then$$fleft(sqrtfrac1n^2+frac1n^4,frac1nright)=left(left(frac1n^2+frac1n^4right)^5/2-frac1n^5right)n^8$$and$$lim_ntoinftyleft(left(frac1n^2+frac1n^4right)^5/2-frac1n^5right)n^8=infty.$$

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

José Carlos SantosJosé Carlos Santos

180k24140254

$endgroup$

share|cite|improve this answer

edited 1 hour ago

answered 2 hours ago

José Carlos SantosJosé Carlos Santos

180k24140254

answered 2 hours ago

José Carlos SantosJosé Carlos Santos

180k24140254

answered 2 hours ago

José Carlos SantosJosé Carlos Santos

180k24140254

2

$begingroup$

I think your substitution gives you a hint. The angles $tanphi =pm 1$ are the problematic ones, i.e., the limit along the lines $pm x=y$. You can check that along these lines there's a divergence.

share|cite|improve this answer

edited 39 mins ago

answered 1 hour ago

olaphusolaphus

763

$endgroup$

add a comment | 

2

$begingroup$

I think your substitution gives you a hint. The angles $tanphi =pm 1$ are the problematic ones, i.e., the limit along the lines $pm x=y$. You can check that along these lines there's a divergence.

share|cite|improve this answer

edited 39 mins ago

answered 1 hour ago

olaphusolaphus

763

$endgroup$

add a comment | 

$begingroup$

I think your substitution gives you a hint. The angles $tanphi =pm 1$ are the problematic ones, i.e., the limit along the lines $pm x=y$. You can check that along these lines there's a divergence.

share|cite|improve this answer

edited 39 mins ago

answered 1 hour ago

olaphusolaphus

763

$endgroup$

share|cite|improve this answer

edited 39 mins ago

answered 1 hour ago

olaphusolaphus

763

answered 1 hour ago

olaphusolaphus

763

answered 1 hour ago

olaphusolaphus

763

0

$begingroup$

In the posting, $$ f(r,phi)=fracr(cos^5phi -sin^5phi
)(cos^2phi -sin^2phi)^2 = underbracefracrcos
phi-sin phi_=gunderbracefracsum_i=1^4
sin^iphicos^4-iphi (cos phi+sin phi)^2 _=h$$

When $f(r,0)=r$, so $lim_r f=0$.

Further, $lim_phi rightarrow
pi/4 h $ exists. We have a sequence $phi_n<pi/4,
phi_nrightarrow pi/4$ s.t. $ cos phi_n > sin phi_n$. We let
$$r_n =sqrtcos phi_n - sin phi_n.$$ Hence $$ g(r_n,phi_n) =
frac1sqrt cos phi_n - sin phi_n rightarrow infty$$
so that $f(r_n,phi_n)$ goes to $infty$.

share|cite|improve this answer

answered 22 mins ago

HK LeeHK Lee

14.2k52363

$endgroup$

add a comment | 

0

$begingroup$

In the posting, $$ f(r,phi)=fracr(cos^5phi -sin^5phi
)(cos^2phi -sin^2phi)^2 = underbracefracrcos
phi-sin phi_=gunderbracefracsum_i=1^4
sin^iphicos^4-iphi (cos phi+sin phi)^2 _=h$$

When $f(r,0)=r$, so $lim_r f=0$.

Further, $lim_phi rightarrow
pi/4 h $ exists. We have a sequence $phi_n<pi/4,
phi_nrightarrow pi/4$ s.t. $ cos phi_n > sin phi_n$. We let
$$r_n =sqrtcos phi_n - sin phi_n.$$ Hence $$ g(r_n,phi_n) =
frac1sqrt cos phi_n - sin phi_n rightarrow infty$$
so that $f(r_n,phi_n)$ goes to $infty$.

share|cite|improve this answer

answered 22 mins ago

HK LeeHK Lee

14.2k52363

$endgroup$

add a comment | 

$begingroup$

In the posting, $$ f(r,phi)=fracr(cos^5phi -sin^5phi
)(cos^2phi -sin^2phi)^2 = underbracefracrcos
phi-sin phi_=gunderbracefracsum_i=1^4
sin^iphicos^4-iphi (cos phi+sin phi)^2 _=h$$

When $f(r,0)=r$, so $lim_r f=0$.

Further, $lim_phi rightarrow
pi/4 h $ exists. We have a sequence $phi_n<pi/4,
phi_nrightarrow pi/4$ s.t. $ cos phi_n > sin phi_n$. We let
$$r_n =sqrtcos phi_n - sin phi_n.$$ Hence $$ g(r_n,phi_n) =
frac1sqrt cos phi_n - sin phi_n rightarrow infty$$
so that $f(r_n,phi_n)$ goes to $infty$.

share|cite|improve this answer

answered 22 mins ago

HK LeeHK Lee

14.2k52363

$endgroup$

share|cite|improve this answer

answered 22 mins ago

HK LeeHK Lee

14.2k52363

answered 22 mins ago

HK LeeHK Lee

14.2k52363

answered 22 mins ago

HK LeeHK Lee

14.2k52363