We assume that $f$ is continuous at $a$ and $f(a) > 0$.
My problem is I am not really clear with the question that ask me to prove the existence of $δ > 0$ such that $f(x) > 0$ for all $x$ satisfying $|x − a| < δ$. Normally, we need to use those statement "there exists $δ > 0$ such that $f(x) > 0$ for all $x$ satisfying $|x − a| < δ$" to proof for example that $f$ is continuous at $a$. Can anyone help me out with this problem?
Try setting $ 0 < \varepsilon < f ( a) $ and consider the respective $ \delta > 0$ from the definition of continuity.
We are looking for some $\delta>0$ and we know that $f$ is continuous at $a$... that's the way to go!
By definition of continuity, for any $\varepsilon>0$, there exists $\delta>0$ such that, if $|x-a|<\delta$, then $|f(x)-f(a)|<\varepsilon$. Note that every choice of $\varepsilon$ gives us at least one $\delta$, so by choosing an appropriate $\varepsilon$, we should get a $\delta$ that prove the statement.
Note also that $|x-a|<\delta$ means $f(a)-\varepsilon<f(x)<f(a)+\varepsilon$. So, if we choose $\varepsilon$ such that $f(a)-\varepsilon>0$, we will have $f(x)>0$. Therefore, $\varepsilon=f(a)$ is a suitable choice (but you can take other values, for example $\frac{f(a)}{2}$).
Now, the argument above is kind of a draft. If you want to write it a clean proof, you just need to put the ideas in the right order:
Let $\varepsilon=f(a)>0$. Then, since $f$ is continuous at $a$, there exists $\delta>0$ such that, if $|x-a|<\delta$, then $|f(x)-f(a)|<\varepsilon$.
This means that, for all $x$ satisfying $|x-a|<\delta$, we have $f(a)+\varepsilon>f(x)>f(a)-\varepsilon=0$. $\square$
