Proving $h$ is continuous for $h(x)=8f(x)-3g(x)$ where $f$ and $g$ are continuous 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 $\delta=min\{\delta_1,\delta_2\}$
$|h(x)-h(a)|=|(8f(x)-3g(x))-(8f(a)-3g(a))|=|(8f(x)-8f(a))+(-3g(a)-(-3g(a))| \leq |8f(x)-8f(a)|+|-3g(a)-(-3g(a))|<8|f(x)-f(a)|-3|g(a)-g(a)|<8.\frac{\epsilon}{5} - 3.\frac{\epsilon}{5}=\epsilon$
 A: As an alternate answer, try proving it by proving the following two
points separately:


*

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

*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.
A: You have to be carefull with the use of Triangle Inquality. For example you have:
$$|8f(x)-8f(a)|+|-3g(a)-(-3g(a))|<8|f(x)-f(a)|-3|g(a)-g(a)|$$
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.
A: 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. 
A: Let $\alpha, \beta \in \mathbb{R}$. If $f$ and $g$ are continuous then $f+g$ are continuous, and so are $\alpha f$ and $\beta g$. Combining those two results yields that $$\alpha f(x)+\beta g(x)$$ is continuous. For your particular case just put $\alpha=8$ and $\beta =-3$. Therefore $$h(c)=8f(x)-3g(x)$$ is continuous and you are done :).
