Find the limit and discuss the continuity of the function. lim(x,y)→(3,1)(x2−2y)

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irwchh · Accepted Answer

Rewiew the given function  f(x,y)=(x2−2y)  Objective is to find the limit when (x,y)⇒(3,1) and discuss the continuity of the function.  To find the limit when (x,y)⇒(3,1), solve  =lim(x,y)→(3,1)(x2−2y)  Since the above limit is not forming any indeterminate form, so evaluate the just by substituting the value of x and y.  =(32−2⋅1)  =9−2  =7  To check the continuity of the function  Any function f(x,y) is a continuous point (x0,y0) if lim(x,y)→(x0,y0)f(x,y)=f(x0,y0)  Here  lim(x,y)→(x0,y0)f(x,y)=f(x0,y0)=lim(x,y)→(3,1)x2−2y=7 and f(x0,y0)=f(3,1)=32−2⋅1=7  Since,  lim(x,y)→(x0,y0)f(x,y)=f(x0,y0)=7  So, the given function is continuous at point (3,1).

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

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