Question

# Find the limit and discuss the continuity of the function. lim_{(x,y)rightarrow(3,1)}(x^2-2y)

Limits and continuity
Find the limit and discuss the continuity of the function. $$\lim_{(x,y)\rightarrow(3,1)}(x^2-2y)$$

2021-01-05
Rewiew the given function
$$f(x,y)=(x^2-2y)$$
Objective is to find the limit when $$(x,y)\Rightarrow(3,1)$$ and discuss the continuity of the function.
To find the limit when $$(x,y)\Rightarrow(3,1)$$, solve
$$=\lim_{(x,y)\rightarrow(3,1)}(x^2-2y)$$
Since the above limit is not forming any indeterminate form, so evaluate the just by substituting the value of x and y.
$$=(3^2-2\cdot1)$$
$$=9-2$$
$$=7$$
To check the continuity of the function
Any function f(x,y) is a continuous point $$(x_0,y_0)$$ if $$\lim_{(x,y)\rightarrow(x_0,y_0)}f(x,y)=f(x_0,y_0)$$
Here
$$\lim_{(x,y)\rightarrow(x_0,y_0)}f(x,y)=f(x_0,y_0)=\lim_{(x,y)\rightarrow(3,1)}x^2-2y=7$$ and $$f(x_0,y_0)=f(3,1)=3^2-2\cdot1=7$$
Since,
$$\lim_{(x,y)\rightarrow(x_0,y_0)} f(x,y)=f(x_0,y_0)=7$$
So, the given function is continuous at point (3,1).