How can I prove that $$\lim_{(x,y)\to(1,1)} \frac {x}{y}=1$$ By epsilon delta?
I am trying and I am stuck:
Proof: Suppose $\epsilon >0$ we want to construct $\delta = \delta (\epsilon ) $ such that $|x/y-1|<\epsilon $ whenever $|x-1|<\delta$ and $|y-1|<\delta$.
But $\vert \frac{x}{y}-1 \vert =\vert \frac {x-y}{y}\vert\leq\frac {|x-1|+|y-1|}{|y|}<\frac {2\delta}{|y-1|}<\frac {2\delta}{\delta}=2$
On
This should work now ;).
$\forall \varepsilon > 0$ find $\delta = \delta(\varepsilon) > 0,$ such that $||(x,y)^T - (1,1)^T||_\infty = \max\{|x - 1|, |y - 1|\} < \delta$ (or $|x - 1| < \delta$ and $|y - 1| < \delta$ implies $f(x,y) = \big|\frac{x}{y} - 1\big| < \varepsilon.$ (you could also use the norm $||\cdot||_1, ||\cdot||_2,$ since alle norms in $\mathbb{R}^2$ are equivalent). Then you get:
$$\Big|\frac{x}{y} - 1\Big| \leq \frac{|x - 1| + |y - 1|}{|y|} < \frac{2\delta}{|y|} < \varepsilon,$$
where the last inequality is equivalent to
$$ 2\delta < |y|\varepsilon \;\;\;\Longrightarrow\;\;\; \delta < \frac{(1+\delta)\varepsilon}{2},$$
since $|y - 1| < \delta \Rightarrow y \in (1 -\delta, 1 + \delta) \Rightarrow |y| < 1 + \delta.$ Further, without loss of generality let's assume $\varepsilon < 2,$ then
$$ \delta < \frac{(1+\delta)\varepsilon}{2} \;\Longleftrightarrow\; \delta - \frac{\delta\varepsilon}{2} < \frac{\varepsilon}{2} \;\Longleftrightarrow\; \delta\Big(1 - \frac{\varepsilon}{2}\Big) < \frac{\varepsilon}{2} \;\Longleftrightarrow\; \delta < \frac{\varepsilon}{2 - \varepsilon}.$$
What you want is to choose $\delta = \min\{ \frac{\epsilon}{4},\frac{1}{2}\}$ and then you know the following.
$$|\frac{x}{y} -1| = |\frac{x-y}{y}| \leq \frac{|x-1|+|y-1|}{|y|} < \frac{2\delta}{|y|} < 4\delta < \epsilon$$
The only thing I changed was that because we know that $\delta \leq \frac{1}{2}$ then $y > \frac{1}{2}$ or $\frac{1}{|y|} < 2$