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Maple determines the bivariate limit at the origin to be 1.
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Context Panel: Assign Function


$f\left(x\,y\right)\=\frac{{x}^{4}\+{x}^{2}\+x{y}^{2}plus;{y}^{2}}{{x}^{2}{x}^{2}yplus;{y}^{2}}$$\stackrel{\text{assign as function}}{\to}$${f}$

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Context Panel: Evaluate and Display Inline

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Context Panel: Limit (Bivariate)


$f\left(x\,y\right)$ = $\frac{{{x}}^{{4}}{+}{x}{}{{y}}^{{2}}{+}{{x}}^{{2}}{+}{{y}}^{{2}}}{{}{{x}}^{{2}}{}{y}{+}{{x}}^{{2}}{+}{{y}}^{{2}}}$$\stackrel{\text{bivariate limit}}{\to}$${1}$



To see why the bivariate limit at the origin might have the value 1, divide the numerator and denominator of $f$ to put $f\left(x\,y\right)$ into the form
$\frac{1\+\frac{{x}^{4}\+x{y}^{2}}{{x}^{2}plus;{y}^{2}}}{1\frac{{x}^{2}y}{{x}^{2}plus;{y}^{2}}}$
The rational functions in the numerator and denominator of this form of $f$ both tend to zero as $\left(x\,y\right)\to \left(0\,0\right)$. Indeed, Maple provides the following bivariate limits, obtained either through the Context Panel, or by application of the limit command with the appropriate syntax.
$\frac{{x}^{4}\+x{y}^{2}}{{x}^{2}plus;{y}^{2}}$$\stackrel{\text{bivariate limit}}{\to}$${0}$

$\frac{{x}^{2}y}{{x}^{2}\+{y}^{2}}$$\stackrel{\text{bivariate limit}}{\to}$${0}$

$\mathrm{limit}\left(\frac{{x}^{4}\+x{y}^{2}}{{x}^{2}plus;{y}^{2}}comma;\left\{xequals;0comma;yequals;0\right\}\right)$ = ${0}$${}$

$\mathrm{limit}\left(\frac{{x}^{2}y}{{x}^{2}\+{y}^{2}}\,\left\{x\=0\,y\=0\right\}\right)$ = ${0}$${}$

Auxiliary Limit (1)

Auxiliary Limit (2)



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Clearly, then, the bivariate limit of $f$ as $\left(x\,y\right)\to \left(0\,0\right)$ will necessarily be 1, in which case the required extension is
$g\left(x\,y\right)\=\{\begin{array}{cc}f\left(x\,y\right)& \left(x\,y\right)\ne \left(0\,0\right)\\ 1& \left(x\,y\right)\=\left(0\,0\right)\end{array}$
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Analytic justification for Auxiliary Limit (1) is based on the following estimate.
$\left{x}^{4}plus;x{y}^{2}\right$

$\le {x}^{4}\+\leftx\right{y}^{2}$

Inequality 3, Table 3.2.1

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$\le {\left({x}^{2}\+{y}^{2}\right)}^{2}\+\sqrt{{x}^{2}\+{y}^{2}}\left({x}^{2}\+{y}^{2}\right)$

Inequalities 4, 6, and 7, Table 3.2.1



Consequently, $\left\frac{{x}^{4}\+x{y}^{2}}{{x}^{2}plus;{y}^{2}}\right\le {x}^{2}plus;{y}^{2}plus;\sqrt{{x}^{2}plus;{y}^{2}}$, and the limiting value of zero is established.
Analytic justification for Auxiliary Limit (2) is based on the following estimate.
$\left{x}^{2}y\right$

$\le {x}^{2}\lefty\right$

Inequality 3, Table 3.2.1

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$\le \left({x}^{2}\+{y}^{2}\right)\sqrt{{x}^{2}\+{y}^{2}}$

Inequalities 5 and 6, Table 3.2.1



Consequently, $\left\frac{{x}^{2}y}{{x}^{2}\+{y}^{2}}\right\le \sqrt{{x}^{2}\+{y}^{2}}$, and the limiting value of zero is established.
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Indeed, $\mathrm{limit}\left(f\left(x\,y\right)\,\left\{x\=0\,y\=0\right\}\right)$ = ${1}$.