# Continuity

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Let $$f$$ be a real function defined on an open interval around point $$a$$.

If $$\displaystyle \lim_{x\to a}f(x) =f(a)$$, then $$f$$ is called continuous at $$a$$. If not, then $$f$$ is called discontinuous at $$a$$.

If $$f$$ is continuous at every point of an open interval $$(c.d)$$, then $$f$$ is called continuous on $$(c.d)$$.

In other words, a function is continuous if the graph can be drawn without lifting the pen from the paper.

Let’s practice!

Consider the function $$f$$ given by

$$\displaystyle f(x)=\begin{cases} 8 x+9&\text{if }x\lt1\\ 9 x^2+8&\text{if }x\geq1\end{cases}$$

Is this function continuous at $$1$$?

The function $$f$$ is left continuous at $$x = 1$$:

$$\begin{array}{rcl} \displaystyle\lim_{x\uparrow1}f(x)&=&\lim_{x\uparrow1}( 8x+9)=8+9=17=f(1)\end{array}$$

The function $$f$$ is right continuous at $$x = 1$$:

$$\begin{array}{rcl}\displaystyle\lim_{x\downarrow1}f(x)&=&\lim_{x\downarrow1} (9 x^2+8)=9+8=17=f(1)\end{array}$$

Since $$f$$ is both right continuous and left continuous at $$1$$, it is continuous at $$1$$.

The graph of $$f$$ is drawn below. It has no jump at $$1$$.

Yes, this function is continuous at $$1$$