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Example: Examples of Functions Continuous at a Single Point
(related to Definition: Continuous Functions at Single Real Numbers)
The following are examples of real functions, which are continuous at single points. They exhibit the main ideas of the definition of continuity of a function at a single point:
Example 1:
The function $f:\mathbb R\to \mathbb R$, $f(x):=2x 3$ is continuous at a point $x=5$.
Proof:
* Let $f(x)=2x 3.$
* Take any $\epsilon > 0$ and set $\delta:=\epsilon/2.$
* Select $x$ such that $0 < x 5 < \delta = \epsilon/2.$
* Then $\delta > x 5$ implies $\epsilon > 2x 5 = 2x10 = (2x 3) 7 = f(x) 7.$
* Because $\epsilon$ might have been chosen arbitrarily small, it follows that the limit $$\lim_{x\to 5}f(x)=\lim_{x\to 5}2x 3=7$$ exists (and equals $7$).
* Thus, $f$ is continuous at the point $x=5.$
Example 2
The function $f:\mathbb R\to \mathbb R$, $f(x):=\frac{x+2}{x^2+3x+2}$ is continuous at a point $x=2$. Please note that the function is not defined for this value of $x$ (division by zero!), even though its limit, as $x$ approaches $2$, exists.
Proof:
 Let $f(x)=\frac{x+2}{x^2+3x+2}.$
 Take any $\epsilon > 0$ and set $\delta:=\min(\epsilon/2,1/2).$
 Select $x$ such that $\delta > x (2) > 0.$
 $1/2 \ge \delta > x+2$ implies $1/2 < x + 2 < 1/2$, so $3/2 < x + 1 < 1/2$. Thus $2 > \frac 1{x+1}$ and $\epsilon/2 < \epsilon x+1.$
 $\epsilon /2 \ge \delta > x+2$ implies together with the above result $x+2 < \epsilon / 2 < \epsilon x+1.$
 Thus, for $x$ with $\delta > x(2) > 0$ we have $$\begin{array}{rcl}\epsilon &>& 2x+2\\
&>& \frac 1{x+1}x+2\\
&=&\frac 1{x+1}1+(x+1)\\
&=&\left\frac1{x+1}+1\right\\
&=&\left\frac{(x+2)}{(x+2)(x+1)}+1\right\\
&=&\left\frac{x+2}{x^2 + 3 x + 2}(1)\right\\
&=&f(x)  (1)
\end{array}$$
 Because $\epsilon$ might have been chosen arbitrarily small, it follows that the limit $$\lim_{x\to 2}f(x)=\lim_{x\to 2}\frac{x+2}{x^2 + 3x + 2}=1$$ exists (and equals $1$).
 Thus, $f$ is continuous at the point $x=2.$
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