- #1

ananonanunes

- 18

- 6

- Homework Statement
- Given the continuous function ##g\colon \mathbb{R} \rightarrow \mathbb{R}## so that ##g(1)=7##, consider the function defined in ##\mathbb{R}^{2}_{\backslash(1,0)}## by$$f(x,y)=\frac {g(x)(x-1)^2y}{2(x-1)^4+y^2}.$$

Say whether the limit ##\lim_ {(x,y) \rightarrow (1,0)} {f(x,y)}## exists.

- Relevant Equations
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This is what I did: $$\lim_ {(x,y) \rightarrow (1,0)} {\frac {g(x)(x-1)^2y}{2(x-1)^4+y^2}}=\lim_ {(x,y) \rightarrow (1,0)} {g(x)y\frac {(x-1)^2}{2(x-1)^4+y^2}}$$ I know that ##\lim_ {(x,y) \rightarrow (1,0)} {g(x)y}=0## and that ##\frac {(x-1)^2}{2(x-1)^4+y^2}## is limited because ##0\leq (x-1)^2\leq 2(x-1)^4\leq 2(x-1)^4+y^2\Rightarrow 0 \leq\frac{(x-1)^2}{2(x-1)^4+y^2}\leq1##, so I concluded that the limit exists and it is 0.

I know this is wrong and I can understand that the limit cannot exist because if you calculate the limit for ##y=(x-1)^2## you get a solution different than 0. But I thought that having a function whose limit is 0 times a function that is limited was enough to conclude that the limit of their multiplication was 0. Is this wrong?

I know this is wrong and I can understand that the limit cannot exist because if you calculate the limit for ##y=(x-1)^2## you get a solution different than 0. But I thought that having a function whose limit is 0 times a function that is limited was enough to conclude that the limit of their multiplication was 0. Is this wrong?