Continuous Function of functions Let $f,g :\mathbb{R} \to \mathbb{R}$ be continous functions. Define $h :\mathbb{R} \to \mathbb{R}\ \ \text{by}\ \ h(x) = 5f(x) -2g(x)$. Show directly from the definition that $h$ is continuous.
All the examples which I have met involved some given value $x=c$ to check.
Any help is much appreciated
 A: Let $x_0$ be given and let $L, M$ be the limits of $f, g$ respectively as they approach $x_0$, then for each $\epsilon>0$ there are $\delta,\delta'$ such that
$$|x-x_0|<\delta\implies |f(x)-L|<\epsilon$$
$$|x-x_0|<\delta' \implies |g(x)-M|<\epsilon$$
Then we note
$$|x-x_0|<\min\{\delta,\delta'\}=\delta_0$$
$$\implies |5f(x)-2g(x)-5L+2M|\le 5|f(x)-L|+2|g(x)-M|$$
$$<7\epsilon.$$
A: DISCLAIMER:See this only you have attempted the solution yourself.
Let $\epsilon>0$ and $x_0\in\mathbb{R}$. Since $f,g$ are continuous there exists $\delta_1,\delta_2>0$ such that $|x-x_0|<\delta_1\rightarrow |f(x)-f(x_1)|<\frac{\epsilon}{10}$ and $|x-x_0|<\delta_2\rightarrow |g(x)-g(x_1)|<\frac{\epsilon}{4}$. Thus if we let $\delta=\max({\delta_1,\delta_2})$ we know that forall $\epsilon>0,\exists \delta$ such that $$|x-x_0|<\delta\rightarrow |h(x)-h(x_1)|\leq5|f(x)-f(x_1)|+2|g(x)-g(x_1)|\leq 5\frac{\epsilon}{10}+2\frac{\epsilon}{4}=\epsilon.$$
A: Let $\{x_n\}_{n=1}^{\infty}$ be a arbitrary convergent sequence with limit $x\in \mathbb{R}$. we have,
$$\lim_{n\rightarrow \infty}h(x_n)=\lim_{n\rightarrow \infty}5f(x_n)-\lim_{n\rightarrow \infty}2g(x_n)=5f(x)-2g(x)$$
and the proof
of the problem is complete.
