The limit I'm solving is: $\lim\limits_{x \to 3} \frac{x}{x^2 + 3}$ = ${1\over 4}$

So when I solve I do

$\left\lvert\frac{x}{x^2 + 3} - {1\over 4} \right\rvert $ < $\epsilon$

I end up with

$\left\lvert\ x-3 \right\rvert $ *

$\left\lvert\ x-1 \right\rvert $ / (4 ($\mathrm{x}^2$ + 3) )

I majorize the latter and apparently plug in 4 and 0 for x. My first question is how I end up plugging in 4 and 0 here. (3 / 4(0 + 3))

I end up getting that 0 < x - 3 < 4$\epsilon$

and derive that delta is the minimum of 1 and 4$\epsilon$.

My final question is how I derived that for delta.

Thanks for help


It is a little bit easier to understand if you replace $x$ by $x=3+u$ with $|u|<\delta$ .

$\bigg|\dfrac x{x^2+3}-\dfrac 14\bigg|=\bigg|-\frac 14\dfrac{(x-1)(x-3)}{(x^2+3)}\bigg|=\underbrace{\frac 14|u|}_{A}\times\underbrace{\dfrac{|u+2|}{|u^2+6u+12|}}_{B}$

We end up with two pieces

  • $A$ which is a small as we want

$\qquad\delta=\varepsilon, |u|<\delta\implies 0\le A\le\frac 14\varepsilon$

  • $B$ which is bounded by a constant

$\qquad\delta=1,|u|<\delta\implies \begin{cases}|u+2|\le|u|+2\le 2+1\le 3\\\\u^2+6u+12\ge 6u+12\ge 12-6|u|\ge 12-6\ge 6\end{cases}$

$\qquad\implies 0\le B\le\frac 12$

To regroup both pieces we need to take $\delta=\min(1,\varepsilon)$

And we get $|x-3|<\delta\implies \bigg|\dfrac x{x^2+3}-\dfrac 14\bigg|\le AB\le \frac 18\varepsilon$


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