Question

Use the method of your choice to evaluate the following limits. lim_{(x,y)rightarrow(1,1)}frac{x^2+xy-2y^2}{2x^2-xy-y^2}

Limits and continuity
ANSWERED
asked 2021-01-10
Use the method of your choice to evaluate the following limits.
\(\lim_{(x,y)\rightarrow(1,1)}\frac{x^2+xy-2y^2}{2x^2-xy-y^2}\)

Answers (1)

2021-01-11
Given:
\(\lim_{(x,y)\rightarrow(1,1)}\frac{x^2+xy-2y^2}{2x^2-xy-y^2}\)
To evaluate:
The given limit.
Solution:
Here,
\(\lim_{(x,y)\rightarrow(1,1)}\frac{x^2+xy-2y^2}{2x^2-xy-y^2}\)
On simplifying the equation,
The required equation is,
\(\lim_{(x,y)\rightarrow(1,1)}\frac{x^2+xy-2y^2}{x^2-xy2-y^2}=\lim_{(x,y)\rightarrow(1,1)}\frac{x+2y}{2x+y}\)
Plug in the values (x,y)=(1,1)
\(\lim_{(x,y)\rightarrow(1,1)}\frac{x^2+xy-2y^2}{x^2-xy2-y^2}=\frac{1+2\times1}{2\times1+1}\)
\(\lim_{(x,y)\rightarrow(1,1)}\frac{x^2+xy-2y^2}{x^2-xy2-y^2}=\frac{1+2}{2+1}\)
\(\lim_{(x,y)\rightarrow(1,1)}\frac{x^2+xy-2y^2}{x^2-xy2-y^2}=\frac{3}{3}\)
\(\lim_{(x,y)\rightarrow(1,1)}\frac{x^2+xy-2y^2}{x^2-xy2-y^2}=1\)
Which is required.
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