Prove that $\lim\limits_{x \to 2} \frac{x-1}{x^2}$= $\frac{1}{4}$, using the $\varepsilon$-$\delta$ definition

Question

So I have to prove $\lim\limits_{x \to 2} \frac{x-1}{x^2}$= $\frac{1}{4}$.

So I need to find $0<|x-2|<\delta$, such that $\left\vert\frac{x-1}{x^2}-\frac{1}{4}\right\vert<\epsilon$.

So $\left\vert\frac{x-1}{x^2}-\frac{1}{4}\right\vert=\frac{(x-2)^2}{4x^2}$.

Now I try to set $\delta=1$, so $0<|x-2|<1$ and $4<4x^2<36$.

And so $\frac{(x-2)^2}{4x^2}<\frac{(x-2)^2}{4}$.

And so $\frac{(x-2)^2}{4}<\epsilon$, when $\delta=\sqrt{4\epsilon}$.

Answer 1

Looks good, because you can use $\color{blue}{4<4x^2}<36$ to say: $$|f(x)-L|=\frac{(x-2)^2}{4x^2} \le \frac{(x-2)^2}{4}$$ And since $|x-2| < \delta$, this means: $$\frac{(x-2)^2}{4} < \frac{\delta^2}{4}$$ You want this under $\varepsilon$, so: $$\frac{\delta^2}{4} < \varepsilon \iff \delta < 2\sqrt{\varepsilon}$$ Note that you know require $\delta \le \min\left\{ 1,2\sqrt{\varepsilon} \right\}$.

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Coming back to the '<' vs '=' (see comments as well): if you set $\delta = 1$ along the way and you end up with requiring $\delta = 2\sqrt{\varepsilon}$, note that you need the strongest (i.e. smallest) bound on $\delta$.

If you have a good $\delta$, any $\delta' < \delta$ will work as well; which is why you'll often see $\delta$ taken to be smaller than (or equal to) all the upper bounds you set / need in your proof.

Answer 2

May be you set $\delta$ to 1,and $0<|x-2|<1$, so $ 4 x^2 $ will between 4 and 36, that is $ 1/4 x^2<1/4 $. Then we can get $ (x-2)^2/(4 x^2)< \frac{1}{4} (x-2)^2 $. Set $ \frac{1}{4} (x-2)^2 < \epsilon $, then $ |x-2|<2 \sqrt{ \epsilon} $. Let $ \delta = Min \{ 2\sqrt \epsilon ,1 \} $