Question

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

Limits and continuity
ANSWERED
asked 2021-01-04
Find the limit and discuss the continuity of the function. \(\lim_{(x,y)\rightarrow(3,1)}(x^2-2y)\)

Answers (1)

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).
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