Burhan Hopper

2021-10-22

Limit and Continuity Find the limit (if it exists) and discuss the continuity of the function.

$\underset{(x,y)\to (0,0)}{lim}\frac{{x}^{2}y}{{x}^{4}+{y}^{2}}$

SkladanH

Skilled2021-10-23Added 80 answers

Find the limit (if it exists) and discuss the continuity of the function.

The function is not solved by taking the limits of the numerator and denominator separately and also by dividing.

As Limits of numerator and denominator are both equal to 0.

Take path from y=0

$\underset{(x,y)\to (0,0)}{lim}\frac{{x}^{2}y}{{x}^{4}+{y}^{2}}=\frac{{x}^{2}\left(0\right)}{{x}^{4}+{0}^{2}}$

$=0$

Take path from x=0

$\underset{(x,y)\to (0,0)}{lim}\frac{{x}^{2}y}{{x}^{4}+{y}^{2}}=\frac{{x}^{2}\left({m}^{2}{x}^{2}\right)}{{x}^{4}+{\left({m}^{2}{x}^{2}\right)}^{2}}$

$=\frac{{m}^{2}{x}^{4}}{x+{\left({m}^{2}{x}^{2}\right)}^{2}}$

$=\frac{{m}^{2}{x}^{4}}{{x}^{4}+{m}^{4}{x}^{4}}$

$=\frac{{m}^{2}}{1+{m}^{4}}$

The different value of m limit will be different.

As the value of limit is depending on values of m.

The function is not continuous at (0,0).

Hence, the given limit does not exist at (0,0) and f is discontinuous at (0,0)

The function is not solved by taking the limits of the numerator and denominator separately and also by dividing.

As Limits of numerator and denominator are both equal to 0.

Take path from y=0

Take path from x=0

The different value of m limit will be different.

As the value of limit is depending on values of m.

The function is not continuous at (0,0).

Hence, the given limit does not exist at (0,0) and f is discontinuous at (0,0)

Jeffrey Jordon

Expert2022-08-24Added 2607 answers

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