How to show that the limit of a multivariable function exists using $\epsilon -\delta$ definition

Question

Prove that $\dfrac{x+4y}{1-x+y} \to 3 $ as $(x,y) \to (1,-1) $

I have no problem bounding the numerator. But I got stucked at the denominator. I have no idea what expression to bound it. Can anyone help? Thanks!

Answer 1

Note that $$|\dfrac{x+4y}{1-x+y} - 3|= \frac {|4(x-1)+(y+1)|}{|1-x+y|}$$

If $\delta$ is small enough, we have $$|4(x-1)+(y+1)|\le |4(x-1)|+|(y+1)|<\frac {\epsilon}{2}$$

On the other hand if $\delta< 1/4 $ we have $$|1-x+y|>1/2$$ whenever $|x-1|<\delta $ and $|y+1|<\delta $

Thus if $\delta$ is small enough we have $$ \frac {|4(x-1)+(y+1)|}{|1-x+y|}<\epsilon$$

Answer 2

$$\biggl|\,\frac{x+4y}{1-x+y}-3\,\biggr|=\biggl|\,\frac{4x+y-3}{1-x+y}\,\biggr|.$$ Now suppose $d\bigl((x,y),(1-1)\bigr)<\delta<\frac12$. Observe that this implies both $|x-1|<\delta$ and $|y+1|<\delta$.

• $|4x+y-3|=|4(x-1)-(y+1)|<4|x-1|+|y+1|<5\delta\,$;
• $|1-x+y|>\bigl||y|-|1-x|\bigr|=|y|-|1-x|$ if $\delta$ is small enough. Note that, since $|y+1|<\delta$, $|y|>1-\delta$ and since $|1-x|<\delta $, $-|1-x|>-\delta$. Thus we obtain finally $$|1-x+y|>1-2\delta,\quad\text{and}\quad\biggl|\,\frac{x+4y}{1-x+y}-3\,\biggr|<\frac{5\delta}{1-2\delta}$$ Given $\varepsilon>0$, we simply have to solve for $\delta$:

$$\frac{5\delta}{1-2\delta}<\varepsilon\iff\delta<\frac\varepsilon{5+2\varepsilon} $$ (strictly speaking, we should check that $\delta<\frac12$, but it is clear this condition is fulfilled if $\varepsilon$ is small enough).