# Example: Examples of Functions Continuous at a Single Point

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 $\epsilon > 2|x- 5| = |2x-10| = |(2x- 3)- 7| = |f(x)- 7|.$ * 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 $$\begin{array}{rcl}\epsilon &>& 2|x+2|\\ &>& \frac 1{|x+1|}|x+2|\\ &=&\frac 1{|x+1|}|1+(x+1)|\\ &=&\left|\frac1{x+1}+1\right|\\ &=&\left|\frac{(x+2)}{(x+2)(x+1)}+1\right|\\ &=&\left|\frac{x+2}{x^2 + 3 x + 2}-(-1)\right|\\ &=&|f(x) - (-1)| \end{array}$$
• 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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