Limits¶

The idea of continuity¶

roughly speaking, a function is continuous on an interval if its graph has no breaks jumps or holes in that interval.

Example¶

• The function \(f(x)=3x^3-x^2+2x-1\) is continous on any interval
• The function \(f(x)=1/x\) is not defined at \(x=0\). it is continous on any interval not contating 0

The Numerical Viewpoint: Continuity at a Point¶

In practical word, continuity of a function at a point is important because it means that small errors in the independent variable lead to small errors in the value of the function. Conversely, if a function is not continous at a point, a small error in input measurement can lead to an enourmous error in output.

Example¶

• Supposed that \(f(x)=x^2\) and that we want to compute \(f(\pi)\). Knowing \(f\) is continous tells up that taking \(x=3.14\) should give a good approximation to \(f(\pi)\), and that we can get as accurate an approximation to \(f(\pi)\) as we want by using enough decimals of \(\pi\).

• Investigate the continuity of \(f(x)=x^2\) at \(x=2\).

Values of \(x^2\) near \(x=2\) |\(x\)|\(1.9\)|\(1.99\)|\(1.999\)|\(2.001\)|\(2.01\)|\(2.1\)| |-|-|-|-|-|-|-| |\(x^2\)|\(3.61\)|\(3.96\)|\(3.996\)|\(4.004\)|\(4.04\)|\(4.41\)|

Whiteboard Question¶

\(\(f(x)=\frac{x^2-4}{x-2},x\ne2\)\)¶

\(\(\lim_{x\rightarrow2}\frac{x^2-4}{x-2}=?\)\)¶

Whiteboard Question¶

\(\(\lim_{h\rightarrow0}\frac{(3+h)^2-9}{h}\)\)¶