# 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}

Question
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.

### Relevant Questions

Use the method of your choice to evaluate the following limits.
$$\lim_{(x,y)\rightarrow(0,\pi/2)}\frac{1-\cos xy}{4x^2y^3}$$
Use the method of your choice to evaluate the following limits.
$$\lim_{(x,y)\rightarrow(2,0)}\frac{1-\cos y}{xy^2}$$
Use the method of your choice to evaluate the following limits.
$$\lim_{(x,y)\rightarrow(1,0)}\frac{\sin xy}{xy}$$
Evaluate the following limits.
$$\lim_{(x,y,z)\rightarrow(1,1,1)}\frac{x^{2}+xy-xz-yz}{x-z}$$
Use the method of your choice to evaluate the following limits.
$$\lim_{(x,y)\rightarrow(4,0)}x^2y\ln xy$$
$$\lim_{(x,y)\rightarrow(1,0)}\frac{y\ln y}{x}$$
$$\lim_{(x,y)\rightarrow(0,1)}\frac{y\sin x}{x(y+1)}$$
Evaluate $$\lim_{x \rightarrow \infty} \frac{\sin h x}{e^x}$$
Evaluate the following limits. $$\lim_{(x,y)\rightarrow(2,2)}\frac{y^2-4}{xy-2x}$$
$$\lim_{(x,y)\rightarrow(2,-1)}(xy^8-3x^2y^3)$$