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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 \(1\).

Continuity image 1.png


Yes, this function is continuous at \(1\)

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