## Proposition 6.2.5: Algebra with Continuous Functions

- The identity function
*f(x) = x*is continuous in its domain. - If
*f(x)*and*g(x)*are both continuous at*x = c*, so is*f(x) + g(x)*at*x = c*. - If
*f(x)*and*g(x)*are both continuous at*x = c*, so is*f(x) * g(x)*at*x = c*. - If
*f(x)*and*g(x)*are both continuous at*x = c*, and*g(x) # 0*, then*f(x) / g(x)*is continuous at*x = c*. - If
*f(x)*is continuous at*x = c*, and*g(x)*is continuous at*x = f(c)*, then the composition*g(f(x))*is continuous at*x = c*.

### Proof:

Suppose *f(x) = x*. Then, given any * > 0*
choose * = / 2*. Then, if

it implies that| x - c | <

. Hence, the identity function is indeed continuous. Was it really necessary to take| f(x) - f(c) | = | x - c | < = / 2 <

*= / 2*?

The sum of continuous functions is continuous follows directly
from the triangle inequality. Take any * > 0*.
There exists * _{1} > 0* such that whenever

we know that| x - c | <_{1}

(because| f(x) - f(c) | <

*f*is continuous at

*c*). There also exists

*such that whenever*

_{2}> 0we know that| x - c | <_{2}

(because| g(x) - g(c) | <

*g*is continuous at

*c*). But then, if we let

*= min(*, we have: if

_{1},_{2})then| x - c | <

That finishes the proof. (That we don't get a simple should not bother us any more).| (f(x) + g(x)) - (f(c) + g(c)) | ≤

| f(x) - f(c) | + | g(x) - g(c) | <

+ = 2

The product of two continuous functions is again continuous, which follows from a simple trick. We will only look at the trick involved, and leave the details to the reader:

With this trick the rest of the proof should not be too difficult.| f(x) g(x) - f(c) g(c) |

= | f(x) g(x) - f(x) g(c) + f(x) g(c) - f(c) g(c) |

≤ | f(x) | | g(x) - g(c) | + | g(c) | | f(x) - f(c) |

A similar trick works for the quotient. Here is the idea:

Can you see how to continue ? Adding and subtracting will help again.| f(x) / g(x) - f(c) / g(c) | = | 1 / g(x) g(c) | | f(x) g(c) - f(c) g(x) |

As for composition of functions, we have to proceed somewhat different: We know that
*f(x)* is continuous at *c*, and *g(x)* is continuous at
*f(c)*. Therefore, given any * > 0* there exists
* _{1} > 0* such that whenever

then| t - d | <_{1}

There also exists| g(t) - g(d) | <

*such that if*

_{2}> 0then| x - c | <_{2}

. (Note that we have replaced the usual by| f(x) - f(c) | <_{1}

*here) Now let*

_{1}and substitute= min(_{1},_{2})

*t = f(x)*and

*d = f(c)*. We have: if

then| x - c | <

and then| f(x) - f(c) | <_{1}

In other words,| g(f(x)) - g(f(c)) | <

*f(g(x))*is continuous at

*x = c*.