Vtyjkyuk

Theorem:

If $f$ is continuous in $ain D_f$ en $f(a)ne 0$, then $f$ contains her sign in an environment around a.

Proof:

$underline f(a)gt 0:$
Take $epsilon =dfracf(a)2$ then it guarantees the existence of a $deltagt 0$ so that

$$|x-a|ltdeltaRightarrow f(a)-f(x)le |f(a)-f(x)|le dfracf(a)2qquad(1.1)
$$
$$Longrightarrow |x-a|ltdeltaRightarrow 0ltdfracf(a)2lt f(x)qquad (1.2) $$

My following questions are:

• Why does it say in step 1.2 f(a) -f(x) and not f(x)-f(a)? Because I'd think I'd take it over from the $epsilon-delta$-definition.




• And is it correct to explain the step 1.1 to 1.2 as because $|f(a) -f(x)|= -f(a)+f(x)$ and $ f(a) -f(x)$ so that the second one cancels the left side of the inequality at step 1.1?

P.S.:

I know I left the second case $f(a)lt 0$ out of it but it's analoguous to the former case. The only thing you have to replace is that $epsilon=dfrac-f(a) 2$ and alter the inequality.

By using the definition of the absolute value, $(1.1)$ could be rewritten as :

$$ |x-a| <delta Rightarrow f(x)in left[ fracf(a)2 , frac3,f(a)2right]$$

My advice would just be to forget $ f(a) -f(x)$ and just work with $|f(a) -f(x)|$

However, you cannot assume you know the sign of $f(a)-f(x)$ as supposed in yous second point. You have to work with the absolute value instead.

That would allow you to write $-epsilon<f(a) -f(x)<epsilon$ or $-epsilon<f(x) -f(a)<epsilon$ (your choice) in order to prove $(1.2)$

If $|u-v|< epsilon$, then $u-v< epsilon$ and $v-u< epsilon$.

Can you take it from here ?