In this problem we show that the function f(x,y)=5x^2-y^2/x^2+y^2 does not have

Question

In this problem we show that the function f(x,y)=5x^2-y^2/x^2+y^2 does not have a limit as (x,y) right arrow (0,0). Suppose that we consider (x,y) right arrow (0,0) along the curve y=2x. Find the limit in this case: Now consider (x, y) right arrow (0,0) along the curve y=3x. Find the limit in this case: Note that the results from (a) and (b) indicate that f has no limit as (x, y) right arrow (0,0)(be sure you can explain why!) To show this more generally, consider (x, y) right arrow (0,0) along the curve y=mx, for arbitrary m. Find the limit in this case:

Explanation / Answer

a)
f(x,y) = (5x^2-y^2)/(x^2+y^2)
f(x,2x) = (5x^2-(2x)^2)/(x^2+(2x)^2)
              = (5x^2-4x^2)/(x^2+4x^2)
              = (x^2)/(5x^2)
              =1/5
Answer: 1/5

b)
f(x,y) = (5x^2-y^2)/(x^2+y^2)
f(x,3x) = (5x^2-(3x)^2)/(x^2+(3x)^2)
              = (5x^2-9x^2)/(x^2+9x^2)
              = (-4 x^2)/(10x^2)
              =-4/10
              = -2/5
Answer: -2/5

c)
since limit comes different , limit DNE

f(x,y) = (5x^2-y^2)/(x^2+y^2)
f(x,mx) = (5x^2-(mx)^2)/(x^2+(mx)^2)
              = (5x^2-m^2 x^2)/(x^2+m^2 x^2)
              = x^2 (5-m^2) / {x^2 (1+m^2)}
              = (5-m^2)/(1+m^2)
Answer:               = (5-m^2)/(1+m^2)

It depoends on m . So Limit DNE

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