Determine if a function is continuous in points












0












$begingroup$


Determine if a function $f(x, y)$ is continuous in points (1, 1) and (0, 0).
$$f(x, y) = begin{cases} frac{x^2+2xy-3y^2}{x^3-y^3}, & x neq y \ A, &x=y=0 \ B, & x =y= 1end{cases}$$



I do struggle with conclusions. How to approach it? I suppose, now one should find the iterated limits, if multivariable limit and iterated limits are equal in the point $M(x_0, y_0)$, therefore, the function is continuous in $M$ (it requires clarification, as well. I am not sure if it's viable to state so).
$$1) lim_{yto1}lim_{xto1}frac{x^2+2xy-3y^2}{x^3-y^3} = lim_{yto1}frac{1+2y-3y^2}{1-y^3} = lim_{yto1}frac{3(y-1)(y+frac{1}{3})}{(y-1)(y^2+y+1)} = frac{4}{3}$$
$$2) lim_{xto1}lim_{yto1}frac{x^2+2xy-3y^2}{x^3-y^3} = lim_{xto1}frac{x^2+2x-3}{x^3-1} = lim_{xto1}frac{(x-1)(x+3)}{(x-1)(x^2+x+1)} = frac{4}{3} $$
Proceed:
$$3) lim_{(x, y) to (1, 1)} frac{x^2+2xy-3y^2}{x^3-y^3} = [y = kx] = lim_{xto1}frac{1+2k-3k^2}{x(1-k^3)} Rightarrow frac{3k^2-2k-1}{k^3-1} = frac{3(k-1)(k+frac{1}{3})}{(k-1)(k^2+k+1) }=[k = 2] = frac{7}{7} = 1 neq frac{4}{3}$$



Now, for $A(0, 0)$:
$$lim_{(x, y)to(0, 0)}frac{x^2+2xy-3y^2}{x^3-y^3} = [y = 2x] = lim_{xto0}frac{-7x^2}{-7x^3} = infty$$



$f(x,y)$ is not continuous in A.






limits multivariable-calculus continuity







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Step 3) is incorrect. $(x,y)$ cannot approach $(1,1)$ along the line $y=2x$
    $endgroup$
    – saulspatz
    Dec 8 '18 at 14:30










  • $begingroup$
    @saulspatz Can it approach $y = x$, then?
    $endgroup$
    – Inter Veridium
    Dec 8 '18 at 14:41






  • 1




    $begingroup$
    No the function isn't defined when $y=x$ except at the special points $(0,0)$ and $(1,1)$.
    $endgroup$
    – saulspatz
    Dec 8 '18 at 14:53
















0












$begingroup$


Determine if a function $f(x, y)$ is continuous in points (1, 1) and (0, 0).
$$f(x, y) = begin{cases} frac{x^2+2xy-3y^2}{x^3-y^3}, & x neq y \ A, &x=y=0 \ B, & x =y= 1end{cases}$$



I do struggle with conclusions. How to approach it? I suppose, now one should find the iterated limits, if multivariable limit and iterated limits are equal in the point $M(x_0, y_0)$, therefore, the function is continuous in $M$ (it requires clarification, as well. I am not sure if it's viable to state so).
$$1) lim_{yto1}lim_{xto1}frac{x^2+2xy-3y^2}{x^3-y^3} = lim_{yto1}frac{1+2y-3y^2}{1-y^3} = lim_{yto1}frac{3(y-1)(y+frac{1}{3})}{(y-1)(y^2+y+1)} = frac{4}{3}$$
$$2) lim_{xto1}lim_{yto1}frac{x^2+2xy-3y^2}{x^3-y^3} = lim_{xto1}frac{x^2+2x-3}{x^3-1} = lim_{xto1}frac{(x-1)(x+3)}{(x-1)(x^2+x+1)} = frac{4}{3} $$
Proceed:
$$3) lim_{(x, y) to (1, 1)} frac{x^2+2xy-3y^2}{x^3-y^3} = [y = kx] = lim_{xto1}frac{1+2k-3k^2}{x(1-k^3)} Rightarrow frac{3k^2-2k-1}{k^3-1} = frac{3(k-1)(k+frac{1}{3})}{(k-1)(k^2+k+1) }=[k = 2] = frac{7}{7} = 1 neq frac{4}{3}$$



Now, for $A(0, 0)$:
$$lim_{(x, y)to(0, 0)}frac{x^2+2xy-3y^2}{x^3-y^3} = [y = 2x] = lim_{xto0}frac{-7x^2}{-7x^3} = infty$$



$f(x,y)$ is not continuous in A.






limits multivariable-calculus continuity







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Step 3) is incorrect. $(x,y)$ cannot approach $(1,1)$ along the line $y=2x$
    $endgroup$
    – saulspatz
    Dec 8 '18 at 14:30










  • $begingroup$
    @saulspatz Can it approach $y = x$, then?
    $endgroup$
    – Inter Veridium
    Dec 8 '18 at 14:41






  • 1




    $begingroup$
    No the function isn't defined when $y=x$ except at the special points $(0,0)$ and $(1,1)$.
    $endgroup$
    – saulspatz
    Dec 8 '18 at 14:53














0












0








0


1



$begingroup$


Determine if a function $f(x, y)$ is continuous in points (1, 1) and (0, 0).
$$f(x, y) = begin{cases} frac{x^2+2xy-3y^2}{x^3-y^3}, & x neq y \ A, &x=y=0 \ B, & x =y= 1end{cases}$$



I do struggle with conclusions. How to approach it? I suppose, now one should find the iterated limits, if multivariable limit and iterated limits are equal in the point $M(x_0, y_0)$, therefore, the function is continuous in $M$ (it requires clarification, as well. I am not sure if it's viable to state so).
$$1) lim_{yto1}lim_{xto1}frac{x^2+2xy-3y^2}{x^3-y^3} = lim_{yto1}frac{1+2y-3y^2}{1-y^3} = lim_{yto1}frac{3(y-1)(y+frac{1}{3})}{(y-1)(y^2+y+1)} = frac{4}{3}$$
$$2) lim_{xto1}lim_{yto1}frac{x^2+2xy-3y^2}{x^3-y^3} = lim_{xto1}frac{x^2+2x-3}{x^3-1} = lim_{xto1}frac{(x-1)(x+3)}{(x-1)(x^2+x+1)} = frac{4}{3} $$
Proceed:
$$3) lim_{(x, y) to (1, 1)} frac{x^2+2xy-3y^2}{x^3-y^3} = [y = kx] = lim_{xto1}frac{1+2k-3k^2}{x(1-k^3)} Rightarrow frac{3k^2-2k-1}{k^3-1} = frac{3(k-1)(k+frac{1}{3})}{(k-1)(k^2+k+1) }=[k = 2] = frac{7}{7} = 1 neq frac{4}{3}$$



Now, for $A(0, 0)$:
$$lim_{(x, y)to(0, 0)}frac{x^2+2xy-3y^2}{x^3-y^3} = [y = 2x] = lim_{xto0}frac{-7x^2}{-7x^3} = infty$$



$f(x,y)$ is not continuous in A.






limits multivariable-calculus continuity







share|cite|improve this question











$endgroup$




Determine if a function $f(x, y)$ is continuous in points (1, 1) and (0, 0).
$$f(x, y) = begin{cases} frac{x^2+2xy-3y^2}{x^3-y^3}, & x neq y \ A, &x=y=0 \ B, & x =y= 1end{cases}$$



I do struggle with conclusions. How to approach it? I suppose, now one should find the iterated limits, if multivariable limit and iterated limits are equal in the point $M(x_0, y_0)$, therefore, the function is continuous in $M$ (it requires clarification, as well. I am not sure if it's viable to state so).
$$1) lim_{yto1}lim_{xto1}frac{x^2+2xy-3y^2}{x^3-y^3} = lim_{yto1}frac{1+2y-3y^2}{1-y^3} = lim_{yto1}frac{3(y-1)(y+frac{1}{3})}{(y-1)(y^2+y+1)} = frac{4}{3}$$
$$2) lim_{xto1}lim_{yto1}frac{x^2+2xy-3y^2}{x^3-y^3} = lim_{xto1}frac{x^2+2x-3}{x^3-1} = lim_{xto1}frac{(x-1)(x+3)}{(x-1)(x^2+x+1)} = frac{4}{3} $$
Proceed:
$$3) lim_{(x, y) to (1, 1)} frac{x^2+2xy-3y^2}{x^3-y^3} = [y = kx] = lim_{xto1}frac{1+2k-3k^2}{x(1-k^3)} Rightarrow frac{3k^2-2k-1}{k^3-1} = frac{3(k-1)(k+frac{1}{3})}{(k-1)(k^2+k+1) }=[k = 2] = frac{7}{7} = 1 neq frac{4}{3}$$



Now, for $A(0, 0)$:
$$lim_{(x, y)to(0, 0)}frac{x^2+2xy-3y^2}{x^3-y^3} = [y = 2x] = lim_{xto0}frac{-7x^2}{-7x^3} = infty$$



$f(x,y)$ is not continuous in A.






limits multivariable-calculus continuity




limits multivariable-calculus continuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 14:38







Inter Veridium

















asked Dec 8 '18 at 14:24









Inter VeridiumInter Veridium

216




216








  • 1




    $begingroup$
    Step 3) is incorrect. $(x,y)$ cannot approach $(1,1)$ along the line $y=2x$
    $endgroup$
    – saulspatz
    Dec 8 '18 at 14:30










  • $begingroup$
    @saulspatz Can it approach $y = x$, then?
    $endgroup$
    – Inter Veridium
    Dec 8 '18 at 14:41






  • 1




    $begingroup$
    No the function isn't defined when $y=x$ except at the special points $(0,0)$ and $(1,1)$.
    $endgroup$
    – saulspatz
    Dec 8 '18 at 14:53














  • 1




    $begingroup$
    Step 3) is incorrect. $(x,y)$ cannot approach $(1,1)$ along the line $y=2x$
    $endgroup$
    – saulspatz
    Dec 8 '18 at 14:30










  • $begingroup$
    @saulspatz Can it approach $y = x$, then?
    $endgroup$
    – Inter Veridium
    Dec 8 '18 at 14:41






  • 1




    $begingroup$
    No the function isn't defined when $y=x$ except at the special points $(0,0)$ and $(1,1)$.
    $endgroup$
    – saulspatz
    Dec 8 '18 at 14:53








1




1




$begingroup$
Step 3) is incorrect. $(x,y)$ cannot approach $(1,1)$ along the line $y=2x$
$endgroup$
– saulspatz
Dec 8 '18 at 14:30




$begingroup$
Step 3) is incorrect. $(x,y)$ cannot approach $(1,1)$ along the line $y=2x$
$endgroup$
– saulspatz
Dec 8 '18 at 14:30












$begingroup$
@saulspatz Can it approach $y = x$, then?
$endgroup$
– Inter Veridium
Dec 8 '18 at 14:41




$begingroup$
@saulspatz Can it approach $y = x$, then?
$endgroup$
– Inter Veridium
Dec 8 '18 at 14:41




1




1




$begingroup$
No the function isn't defined when $y=x$ except at the special points $(0,0)$ and $(1,1)$.
$endgroup$
– saulspatz
Dec 8 '18 at 14:53




$begingroup$
No the function isn't defined when $y=x$ except at the special points $(0,0)$ and $(1,1)$.
$endgroup$
– saulspatz
Dec 8 '18 at 14:53










1 Answer
1






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$begingroup$

Note that when $y=x$ the numerator vanishes as well as the denominator, so you should simplify by takin a factor of $x-y$ out of both: $$
frac{x^2+2xy-3y^2}{x^3-y^3} ={(x-y)(x+3y)over(x-y)(x^2+xy+y^2)}$$



Now it's easy to see that $f$ is continuous at $(1,1)$ provided that $B=frac43$ and not otherwise.



Your argument at $(0,0)$ is correct.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Does the equality of iterated and multivariable limits assume differentiability?
    $endgroup$
    – Inter Veridium
    Dec 8 '18 at 17:39










  • $begingroup$
    No, we're just talking about continuity. Differentiability plays no part. You just have the quotient of two continuous functions. You don't need this business of iterated limits at all.
    $endgroup$
    – saulspatz
    Dec 8 '18 at 17:41











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1 Answer
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active

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2












$begingroup$

Note that when $y=x$ the numerator vanishes as well as the denominator, so you should simplify by takin a factor of $x-y$ out of both: $$
frac{x^2+2xy-3y^2}{x^3-y^3} ={(x-y)(x+3y)over(x-y)(x^2+xy+y^2)}$$



Now it's easy to see that $f$ is continuous at $(1,1)$ provided that $B=frac43$ and not otherwise.



Your argument at $(0,0)$ is correct.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Does the equality of iterated and multivariable limits assume differentiability?
    $endgroup$
    – Inter Veridium
    Dec 8 '18 at 17:39










  • $begingroup$
    No, we're just talking about continuity. Differentiability plays no part. You just have the quotient of two continuous functions. You don't need this business of iterated limits at all.
    $endgroup$
    – saulspatz
    Dec 8 '18 at 17:41
















2












$begingroup$

Note that when $y=x$ the numerator vanishes as well as the denominator, so you should simplify by takin a factor of $x-y$ out of both: $$
frac{x^2+2xy-3y^2}{x^3-y^3} ={(x-y)(x+3y)over(x-y)(x^2+xy+y^2)}$$



Now it's easy to see that $f$ is continuous at $(1,1)$ provided that $B=frac43$ and not otherwise.



Your argument at $(0,0)$ is correct.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Does the equality of iterated and multivariable limits assume differentiability?
    $endgroup$
    – Inter Veridium
    Dec 8 '18 at 17:39










  • $begingroup$
    No, we're just talking about continuity. Differentiability plays no part. You just have the quotient of two continuous functions. You don't need this business of iterated limits at all.
    $endgroup$
    – saulspatz
    Dec 8 '18 at 17:41














2












2








2





$begingroup$

Note that when $y=x$ the numerator vanishes as well as the denominator, so you should simplify by takin a factor of $x-y$ out of both: $$
frac{x^2+2xy-3y^2}{x^3-y^3} ={(x-y)(x+3y)over(x-y)(x^2+xy+y^2)}$$



Now it's easy to see that $f$ is continuous at $(1,1)$ provided that $B=frac43$ and not otherwise.



Your argument at $(0,0)$ is correct.






share|cite|improve this answer









$endgroup$



Note that when $y=x$ the numerator vanishes as well as the denominator, so you should simplify by takin a factor of $x-y$ out of both: $$
frac{x^2+2xy-3y^2}{x^3-y^3} ={(x-y)(x+3y)over(x-y)(x^2+xy+y^2)}$$



Now it's easy to see that $f$ is continuous at $(1,1)$ provided that $B=frac43$ and not otherwise.



Your argument at $(0,0)$ is correct.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 8 '18 at 15:05









saulspatzsaulspatz

14.5k21329




14.5k21329












  • $begingroup$
    Does the equality of iterated and multivariable limits assume differentiability?
    $endgroup$
    – Inter Veridium
    Dec 8 '18 at 17:39










  • $begingroup$
    No, we're just talking about continuity. Differentiability plays no part. You just have the quotient of two continuous functions. You don't need this business of iterated limits at all.
    $endgroup$
    – saulspatz
    Dec 8 '18 at 17:41


















  • $begingroup$
    Does the equality of iterated and multivariable limits assume differentiability?
    $endgroup$
    – Inter Veridium
    Dec 8 '18 at 17:39










  • $begingroup$
    No, we're just talking about continuity. Differentiability plays no part. You just have the quotient of two continuous functions. You don't need this business of iterated limits at all.
    $endgroup$
    – saulspatz
    Dec 8 '18 at 17:41
















$begingroup$
Does the equality of iterated and multivariable limits assume differentiability?
$endgroup$
– Inter Veridium
Dec 8 '18 at 17:39




$begingroup$
Does the equality of iterated and multivariable limits assume differentiability?
$endgroup$
– Inter Veridium
Dec 8 '18 at 17:39












$begingroup$
No, we're just talking about continuity. Differentiability plays no part. You just have the quotient of two continuous functions. You don't need this business of iterated limits at all.
$endgroup$
– saulspatz
Dec 8 '18 at 17:41




$begingroup$
No, we're just talking about continuity. Differentiability plays no part. You just have the quotient of two continuous functions. You don't need this business of iterated limits at all.
$endgroup$
– saulspatz
Dec 8 '18 at 17:41


















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