Step 1

We have to evaluate the limit of the function with three variable:

\(\lim_{(x,y,z)\rightarrow(1,1,1)}\frac{x^{2}+xy-xz-yz}{x-z}\)

After putting value of limit we get that it is 00 form

We can solve this limit by laws of factorization.

Step 2

Solving by factorization,

\(\lim_{(x,y,z)\rightarrow(1,1,1)}\frac{x^{2}+xy-xz-yz}{x-z}=\lim_{(x,y,z)\rightarrow(1,1,1)}\frac{x^{2}+xy-xz-yz}{x-z}\)

\(=\lim_{(x,y,z)\rightarrow(1,1,1)}\frac{x^{2}-xz+xy-yz}{x-z}\)

\(=\lim_{(x,y,z)\rightarrow(1,1,1)}\frac{x(x-z)+y(x-z)}{x-z}\)

\(=\lim_{(x,y,z)\rightarrow(1,1,1)}\frac{(x-z)(x+y)}{(x-z)}\)

\(=\lim_{(x,y,z)\rightarrow(1,1,1)}(x+y)\)

=1+1

=2

Hence, value of given limit is 2.

We have to evaluate the limit of the function with three variable:

\(\lim_{(x,y,z)\rightarrow(1,1,1)}\frac{x^{2}+xy-xz-yz}{x-z}\)

After putting value of limit we get that it is 00 form

We can solve this limit by laws of factorization.

Step 2

Solving by factorization,

\(\lim_{(x,y,z)\rightarrow(1,1,1)}\frac{x^{2}+xy-xz-yz}{x-z}=\lim_{(x,y,z)\rightarrow(1,1,1)}\frac{x^{2}+xy-xz-yz}{x-z}\)

\(=\lim_{(x,y,z)\rightarrow(1,1,1)}\frac{x^{2}-xz+xy-yz}{x-z}\)

\(=\lim_{(x,y,z)\rightarrow(1,1,1)}\frac{x(x-z)+y(x-z)}{x-z}\)

\(=\lim_{(x,y,z)\rightarrow(1,1,1)}\frac{(x-z)(x+y)}{(x-z)}\)

\(=\lim_{(x,y,z)\rightarrow(1,1,1)}(x+y)\)

=1+1

=2

Hence, value of given limit is 2.