<s>[INST] Prove $\lim_{x\to 2} f(x) = 3$. 
Let $f(x) = \begin{cases}x^2 + 2x - 5;  &x\ne 2\\ 0 &x=2.\end{cases}$. Prove that $\lim\limits_{x\to 2}f(x) = 3$.

By theorem,
$$\lim\limits_{x\to a}f(x) = L \iff \lim_{x\to a+}f(x) = L = \lim_{x\to a-}f(x).$$
Demostración: Let $\epsilon > 0$. Note that $|x^2 +2x - 5 - 3| = |x+4||x-2|$.
Showing right hand limit. Let $0 < x-2 \le 1$. Then $|x+4| = |x-2+6| \leq 1 + 6$. Set $\delta_1 = \min\{1,\frac \epsilon 7\}$. Then if $0 < x-2<\delta_1$, this implies that $|x+4||x-2| \le 7|x-2|<\epsilon$.
Showing left hand limit. Let $-1 < x-2 < 0$. Then $|x+4| \le |x-2| + 4 < 4$. Then set $\delta_2 = \min\{1,\frac \epsilon 4\}$. Then if $-\delta_2 < x-2<0$, this implies $|x+4||x-2|\le 4|x-2|<\epsilon$.
Because the $\lim_{x\to 2+}f(x) = 3 = \lim_{x\to 2-}f(x)$, we conclude that $\lim_{x\to 2}f(x) = 3$.
I am a bit iffy on proving left hand limits. Please help.
 [/INST] Let $0 < |x - 2| < \delta_{\varepsilon}$. Then we have:
\begin{align*}
|f(x) - 3| & = |x^{2} + 2x - 5 - 3|\\\\
& = |x^{2} + 2x - 8|\\\\
& = |(x^{2} - 4x + 4) + (6x - 12)|\\\\
& = |(x-2)^{2} + 6(x-2)|\\\\
& \leq |x - 2|^{2} + 6|x - 2|\\\\
& < \delta^{2}_{\varepsilon} + 6\delta_{\varepsilon} := \varepsilon
\end{align*}
Can you take it from here?
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