<s>[INST] Showing Function is Continuous Let $f: \mathbb{R} \backslash \{2\} \to \mathbb{R}$ be the function given by $f(x) = \frac{2x^2+x-10}{3x-6}$.
Let $g: \mathbb{R} \to \mathbb{R}$ given by:
$$g(x)=\begin{cases}{f(x)} & \text{if } x \neq 2 \\ 3 &\text{if } x = 2  \end{cases}$$
Prove that $g$ is continuous for all $x \in \mathbb{R}$.
Workings:
Suppose $\epsilon > 0$ and let $\delta =$ ______
Then in $|x-2| < \delta$, we have
$|g(x) - g(c)| = \left|\frac{2x^2+x-10}{3x-6} - 3\right|$
$=  \left|\frac{2x+5}{3} - 3\right|$
$= \left|\frac{2x+5}{3} - \frac{9}{3}\right|$
$= \left|\frac{2x-4}{3}\right|$
$= \frac{2}{3}\left|x-2\right|$
$< \frac{2}{3}\delta$
Now I'm not too sure what to do. Any help will be appreciated.
 [/INST] You just have to calculate the 2 limits :
$$l_1 = \lim_{x \rightarrow 2;  x > 2} g(x) \text{ and } l_2 = \lim_{x \rightarrow 2;  x < 2} g(x)$$
and show that $l_1 = l_2 = g(2) = 3$. But $$l_1 = \lim_{x \rightarrow 2;  x > 2} g(x) = \lim_{x \rightarrow 2;  x > 2} f(x) = \lim_{x \rightarrow 2;  x > 2} \frac{2x^2 + x - 10}{3x-6} = \lim_{x \rightarrow 2;  x > 2} \frac{(2x^2 + x - 10)'}{(3x-6)'} = \lim_{x \rightarrow 2;  x > 2} \frac{4x+1}{3} = 3$$ (using L'Hopital's rule). And similarly $l_2 = 3$. So the function is continuous.
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