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

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.