Calculating the Limit of a Function that divisor approaches to zero I try to calculate the limit of a function:
$$ f : \mathbb{R} \to \mathbb{R}, f(x) = \begin{cases}x^2-3 , x \geq 0 \\ 3x, x < 0\end{cases}$$
Evaluate $ \lim_{x\to2} \frac{f(x)-f(2)}{x-2}.$
A)$0\ \ \ \ $ B)$4\ \ \ \ $ C)$2\ \ \ \ $ D)$3\ \ \ \ $ E) No limit

My Deduction
$f(2) = 1$
When $x = 2,$ expression's divisor becomes zero, so I approach $2$ from negative and positive side.
Approaching from negative: ($\epsilon$ is very small positive value)
$x = 2 - \epsilon$
$$  \frac{4-2\epsilon+\epsilon^2 -1}{-\epsilon}   $$
$$  \frac{3-2\epsilon+\epsilon^2}{-\epsilon}   $$
negative infinity
Approaching from positive:
$x = 2 + \epsilon$
$$  \frac{4+2\epsilon+\epsilon^2 -1}{\epsilon}   $$
$$  \frac{3+2\epsilon+\epsilon^2}{\epsilon}   $$
positive infinity
Since approaching from negative and positive has different values, it has no limit, so answer should be E.
Key says B.
Where am I wrong?
 A: Your error is in the numerator, because $f(x) = x^2 - 3$ in this region, the numerator in both cases should be $((2 \pm \epsilon)^2 - 3) - 1 = (\epsilon^2 \pm 4\epsilon + 4) - 4 = \epsilon^2 \pm 4\epsilon.$
So in your left-handed limit you should have $\lim_{\epsilon \to 0^+} \frac{\epsilon^2 - 4\epsilon}{-\epsilon} = \lim_{\epsilon \to 0^+} -\epsilon + 4 = 4,$ and in your right-handed limit you should have $\lim_{\epsilon \to 0^+} \frac{\epsilon^2 + 4\epsilon}{\epsilon} = \lim_{\epsilon \to 0^+} \epsilon + 4 = 4.$ Both one-sided limits go to $4,$ so the limit exists and is $4.$
Of course, as Lone Student points out, this is all a bit unnecessary, you can reach the same conclusion by factoring using the difference between two squares, or alternatively if you're allowed to use derivative rules then you can simply recognize this as $f'(2) = 2(2) = 4.$
A: You don't need $\varepsilon$. Just use limit rules:
As you said $f(2)=1$. Then you can complete:
$$\lim_{x\to 2}\frac{x^2-4}{x-2}=\lim_{x\to 2}(x+2)=4.$$
