I'm trying to prove that $h$ is continuous for $h(x)=8f(x)-3g(x)$ where $f$ and $g$ are continuous. I'm not entirely sure if my choice of $\delta$ for the proof works.

My proof:

For any $\epsilon >0$ then by the continuity of $f$ and $g$ at $a$ let $\delta_1$ and $\delta_2$ be such that $|x-a|<\delta_1 \rightarrow |f(x)-f(a)|<\frac{\epsilon}{5}$, $|x-a|<\delta_2 \rightarrow |g(x)-g(a)|<\frac{\epsilon}{5}$.

Now we need to find a $\delta$ s.t. $|h(x)-h(a)|<\epsilon$. Set $\delta=min\{\delta_1,\delta_2\}$ $|h(x)-h(a)|=|(8f(x)-3g(x))-(8f(a)-3g(a))|=|(8f(x)-8f(a))+(-3g(a)-(-3g(a))| \leq |8f(x)-8f(a)|+|-3g(a)-(-3g(a))|<8|f(x)-f(a)|-3|g(a)-g(a)|<8.\frac{\epsilon}{5} - 3.\frac{\epsilon}{5}=\epsilon$


3 Answers 3


As an alternate answer, try proving it by proving the following two points separately:

  1. The function $c$ defined by $c(x)=a(x)b(x)$ is continuous whenever $a$ and $b$ are continuous.

  2. The function $c$ defined by $c(x)=a(x)+b(x)$ is continuous whenever $a$ and $b$ are continuous.

Then, by point 1, $8f$ and $-3g$ are continuous. By point 2, $h=8f-3g$ is continuous.


You have to be carefull with the use of Triangle Inquality. For example you have:


which is never true. In fact the inequality holds in the opposite direction.

You can use:

$$|(8f(x)-8f(a))+(-3g(a)-(-3g(a))| = |(8f(x) - 8f(a)) + (3g(a)) - 3g(x))| \le |8f(x) - 8f(a)| + |3g(a)) - 3g(x)|$$

This will also cause you changes in how you use $\epsilon$, as you will have to use $\frac{\epsilon}{11}$, instead of $\frac{\epsilon}{5}$.

Outside of this you get the idea spot on.

  • $\begingroup$ Thanks for the great answer but why $|8f(x)-8f(a)|+|-3g(x)-(-3g(a))|<8|f(x)-f(a)|-3|g(x)-g(a)|$ is never true. I assumed it was true because $|8f(x)-8f(a)|+|-3g(x)-(-3g(a))|=8|f(x)-f(a)|+3|g(x)-g(a)|<8|f(x)-f(a)|-3|g(x)-g(a)|$, as $|f(x)-f(a)|$ and $|g(x)-g(a)|$ are always positive. $\endgroup$
    – iza
    Oct 26, 2017 at 20:24
  • 1
    $\begingroup$ @iza Isn't $8|f(x) - f(a)| - 3|g(x)-g(a)| \le 8|f(x) - f(a)| \le 8|f(x) - f(a)| + 3|g(x)-g(a)|$? $\endgroup$
    – Stefan4024
    Oct 26, 2017 at 20:29
  • $\begingroup$ Oh yh, sorry lol. Really stupid mistake. $\endgroup$
    – iza
    Oct 26, 2017 at 20:34

Lets see, it is easier to show that if $f$ and $g$ are continuous at $x:=x_0$ then $\alpha f+\beta g$ is continuous at $x=x_0$.

We first prove that $f+g$ is continuous at $x=x_0$. Let $\varepsilon >0$, we know that there exists $\delta_1,\delta_2>0$ such that $$0<|x-x_0|<\delta_1 \implies |f(x)-f(x_0)|<\frac{\varepsilon}{2}$$ and $$0<|x-x_0|<\delta_2 \implies |g(x)-g(x_0)|<\frac{\varepsilon}{2}.$$ Choosing $\delta:=\min\{ \delta_1,\delta_2 \}$ we get that if $0<|x-x_0|<\delta$ then $|(f+g)(x)-(f+g)(x_0)|=|f(x)-f(x_0)+g(x)-g(x_0)|$ and so $$|(f+g)(x)-(f+g)(x_0)|\leq |f(x)-f(x_0)|+|g(x)-g(x_0)|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$$ Hence $f+g$ is continuous at $x_0$.

Now, let $\epsilon>0$ and let $c \in \mathbb{R}$. If $c=0$ then $cf(x) \equiv 0$ and so $cf$ is continuous. If $c \neq 0$ then if there exists $\delta'>0$ such that $0<|x-x_0|<\delta'$ we get $|cf(x)-cf(x_0)|=|c||f(x)-f(x_0)|=\epsilon |c|$. Since $\epsilon >0$ is arbitrary, the result follows.

And so, if $f$ and $g$ are continuous then $\alpha f$ and $\beta g$ arec ontinuous and therefore $\alpha f+\beta g$ is continuous.

In your example you just take $\alpha=8$ and $\beta=-3$ to conclude that $h$ is continuous.


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