# Evaluate the following limits. lim_{(x,y,z)rightarrow(1,1,1)}frac{x^2+xy-xz-yz}{x-z}

Zoe Oneal 2021-02-05 Answered
Evaluate the following limits.
$\underset{\left(x,y,z\right)\to \left(1,1,1\right)}{lim}\frac{{x}^{2}+xy-xz-yz}{x-z}$
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wornoutwomanC
We have to evaluate the limit of the function with three variable:
$\underset{\left(x,y,z\right)\to \left(1,1,1\right)}{lim}\frac{{x}^{2}+xy-xz-yz}{x-z}$
After putting value of limit we get that it is $\frac{0}{0}$ form
We can solve this limit by laws of factorization.
Solving by factorization,
$\underset{\left(x,y,z\right)\to \left(1,1,1\right)}{lim}\frac{{x}^{2}+xy-xz-yz}{x-z}=\underset{\left(x,y,z\right)\to \left(1,1,1\right)}{lim}\frac{{x}^{2}-xz+xy-yz}{x-z}$
$=\underset{\left(x,y,z\right)\to \left(1,1,1\right)}{lim}\frac{x\left(x-z\right)+y\left(x-z\right)}{x-z}$
$=\underset{\left(x,y,z\right)\to \left(1,1,1\right)}{lim}\frac{\left(x-z\right)\left(x+y\right)}{\left(x-z\right)}$
$=\underset{\left(x,y,z\right)\to \left(1,1,1\right)}{lim}\left(x+y\right)$
$=1+1$
$=2$
Hence, value of given limit is 2.