- #1

- 169

- 0

I'm supposed to use the definition of limit to show that [tex]\lim_{\substack{x\rightarrow 3\\y\rightarrow -1}} f(x,y) = 5[/tex]

where [tex]f(x,y) = x - 2y[/tex]

I split the function into two: [tex]H(x,y) = x[/tex] and [tex]G(x,y) = -2y[/tex]

It can be shown that H + G = f and that the addition of their limits equals the limit of f

So I let [tex]\epsilon > 0[/tex] and set [tex]\delta = \epsilon/2[/tex], so that whenever [tex]\sqrt{(x-3)^2+(y+1)^2} < \delta[/tex] then [tex]|H(x,y) - 3|= |x - 3| < \epsilon/2[/tex] and [tex]|G(x,y) - 2|= |-2y - 2|= 2 |y + 1| < \epsilon/2[/tex]

[tex]|H(x,y) + G(x,y) - (3 + 2)| = |H(x,y) -3 + G(x,y) -2| = |(x-3) + 2(y+1)| <= |x-3| + |2 (y+1)| < 2 \sqrt{(x-3)^2+(y+1)^2} = 2 \delta = \epsilon [/tex]

So, what all is wrong?

where [tex]f(x,y) = x - 2y[/tex]

I split the function into two: [tex]H(x,y) = x[/tex] and [tex]G(x,y) = -2y[/tex]

It can be shown that H + G = f and that the addition of their limits equals the limit of f

So I let [tex]\epsilon > 0[/tex] and set [tex]\delta = \epsilon/2[/tex], so that whenever [tex]\sqrt{(x-3)^2+(y+1)^2} < \delta[/tex] then [tex]|H(x,y) - 3|= |x - 3| < \epsilon/2[/tex] and [tex]|G(x,y) - 2|= |-2y - 2|= 2 |y + 1| < \epsilon/2[/tex]

[tex]|H(x,y) + G(x,y) - (3 + 2)| = |H(x,y) -3 + G(x,y) -2| = |(x-3) + 2(y+1)| <= |x-3| + |2 (y+1)| < 2 \sqrt{(x-3)^2+(y+1)^2} = 2 \delta = \epsilon [/tex]

So, what all is wrong?

Last edited: