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# 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}

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Limits and continuity
asked 2020-12-05
Evaluate the following limits or determine that they do not exist.
$$\lim_{(x,y,z)\rightarrow(2,2,3)}\frac{x^2z-3x^2-y^2z+3y^2}{xz-3x-yz+3y}$$

## Answers (1)

2020-12-06
We have to evaluate the limits:
$$\lim_{(x,y,z)\rightarrow(2,2,3)}\frac{x^2z-3x^2-y^2z+3y^2}{xz-3x-yz+3y}$$
Putting x=2, y=2 and z=3 in the function, we get
$$\lim_{(x,y,z)\rightarrow(2,2,3)}\frac{x^2z-3x^2-y^2z+3y^2}{xz-3x-yz+3y}=\frac{2^2\times3-3\times2^2-2^2\times3+3\times2^2}{2\times3-3\times2-2\times3+3\times2}$$
$$=\frac{12-12-12+12}{6-6-6+6}$$
$$=\frac{0}{0}$$
So it is indeterminate form of $$\frac{0}{0}$$
Solving the limit by factorizing the numerator and denominator,
$$\lim_{(x,y,z)\rightarrow(2,2,3)}\frac{x^2z-3x^2-y^2z+3y^2}{xz-3x-yz+3y}=\lim_{(x,y,z)\rightarrow(2,2,3)}\frac{x^2(z-3)-y^2(z-3)}{x(z-3)-y(z-3)}$$
$$=\lim_{(x,y,z)\rightarrow(2,2,3)}\frac{(z-3)(x^2-y^2)}{(z-3)(x-y)}$$
$$=\lim_{(x,y,z)\rightarrow(2,2,3)}\frac{(x-y)(x+y)}{(x-y)}$$
$$=\lim_{(x,y,z)\rightarrow(2,2,3)}(x+y)$$
$$=2+2$$
$$=4$$
Hence, value of limit is 4.

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