# Evaluate the following limits or determine that they do not exist. lim_{(x,y,z)rightarrow(2,2,3)}frac{x^2z-3x^2-y^2z+3y^2}{xz-3x-yz+3y}

Evaluate the following limits or determine that they do not exist.
$\underset{\left(x,y,z\right)\to \left(2,2,3\right)}{lim}\frac{{x}^{2}z-3{x}^{2}-{y}^{2}z+3{y}^{2}}{xz-3x-yz+3y}$
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StrycharzT
We have to evaluate the limits:
$\underset{\left(x,y,z\right)\to \left(2,2,3\right)}{lim}\frac{{x}^{2}z-3{x}^{2}-{y}^{2}z+3{y}^{2}}{xz-3x-yz+3y}$
Putting x=2, y=2 and z=3 in the function, we get
$\underset{\left(x,y,z\right)\to \left(2,2,3\right)}{lim}\frac{{x}^{2}z-3{x}^{2}-{y}^{2}z+3{y}^{2}}{xz-3x-yz+3y}=\frac{{2}^{2}×3-3×{2}^{2}-{2}^{2}×3+3×{2}^{2}}{2×3-3×2-2×3+3×2}$
$=\frac{12-12-12+12}{6-6-6+6}$
$=\frac{0}{0}$
So it is indeterminate form of $\frac{0}{0}$
Solving the limit by factorizing the numerator and denominator,
$\underset{\left(x,y,z\right)\to \left(2,2,3\right)}{lim}\frac{{x}^{2}z-3{x}^{2}-{y}^{2}z+3{y}^{2}}{xz-3x-yz+3y}=\underset{\left(x,y,z\right)\to \left(2,2,3\right)}{lim}\frac{{x}^{2}\left(z-3\right)-{y}^{2}\left(z-3\right)}{x\left(z-3\right)-y\left(z-3\right)}$
$=\underset{\left(x,y,z\right)\to \left(2,2,3\right)}{lim}\frac{\left(z-3\right)\left({x}^{2}-{y}^{2}\right)}{\left(z-3\right)\left(x-y\right)}$
$=\underset{\left(x,y,z\right)\to \left(2,2,3\right)}{lim}\frac{\left(x-y\right)\left(x+y\right)}{\left(x-y\right)}$
$=\underset{\left(x,y,z\right)\to \left(2,2,3\right)}{lim}\left(x+y\right)$
$=2+2$
$=4$
Hence, value of limit is 4.
Jeffrey Jordon