Let $f(x,y) = 4y$. We know that $\lim_{(x,y) \to (1,2)} f(x,y) = 8$
Find a $\delta > 0 / ||(x,y) - (1,2)||_2 < \delta \Rightarrow |f(x,y) - 8| < 0.1$
I'm trying to find a $ \delta$ in terms of $\epsilon$ like usually and then replace epsilon by 0.1, but I get stuck everytime whenever I try to bring an inequality in
Any tips/useful inequalities would be greatly appreciated
(Also, if you have a table with simple and useful inequalities I'd also be thankful if you shared it. Nothing too complicated, this is an introduction to myltivariable calculus)
You want $|f(x,y) - 8| <.1 \Leftarrow$
$|4y - 8| < .1 \Leftarrow$
$|y - 2| < .025 \Leftarrow$
$||(x,y) - (1,2)|| < .025$.
Trick is to realize the $x$ has no bearing. And $||(x,y) - (1,2)|| < \delta \implies |y-2| < \delta$.
And as the function is simply multiplying by a constant we just need delta to be inversely proportional to the constant.
...
Formally:
If $||(x,y) - (1,2)|| < .025$ then
$\sqrt {(1-x)^2 + (2-y)^2} < 0.025$ and
$\sqrt{(2-y)^2 } \le \sqrt {(1-x)^2 + (2-y)^2} < 0.025$
$|y-2| < 0.025$
$|4y - 8| < 0.1$
$|f(x,y) - 8| < 0.1$.
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2nd go as the above was too slick.
$||(x,y) - (1,2)|| < \delta$
$\sqrt{(y-2)^2 + (1-x)^2} < \delta$
$(y-2)^2 + (1-x)^2 < \delta^2$
$(y-2)^2 < \delta^2 - (1-x)^2 < \delta^2$
$(y-2)^2 < \delta^2$
$|y-2| < \delta$
$|4y - 8| < 4\delta$
$|f(x,y) - 8| < 4\delta$.
We want $4\delta \le .1$.
So $\delta \le .025$ does it.