Understanding the Definition of Continuity I'm a bit confused about what my professor mentioned about the continuity of functions. Let $f$ be continuous on $[a, b]$, and suppose there exists a $c \in [a, b]$ such that $f(c) > 0$. Then there exists a $\delta$ such that $f(x) > 0$ for $|x-c| &lt \delta$.
Why is this the case? (This is not homework).
 A: The definition of $f$ being continuous at a point $c$ is that for all $\epsilon > 0$ there is a $\delta >0$ such that $\lvert x - c &lt \epsilon \Rightarrow \lvert f(x) - f(c) \lvert &lt \epsilon$.
Now let $c\in [a,b]$ be such that $f(c) > 0$. Let $\epsilon = \frac{f(c)}{2} >0$. Then by definition there is a $\delta >0$ such that if $\lvert x - c\lvert &lt \delta$ then $\lvert f(x) - f(c) \lvert &lt \epsilon$. So if $\lvert x - c\lvert &lt \delta$ then
$$\begin{align}
&\lvert f(x) - f(c) \lvert &lt \epsilon \Rightarrow \\
&\lvert f(x) - f(c) \lvert &lt \frac{f(c)}{2} \Rightarrow \\
&-\frac{f(c)}{2} &lt f(x) - f(c) &lt \frac{f(c)}{2} \Rightarrow \\
& 0&lt \frac{f(c)}{2} &lt f(x).
\end{align}
$$ 
A: Ok, I am going to mix and match formality and informality, here. Because I can. We are told from the outset that $f$ is continuous. In particular, this means it is continuous at $c$. What is one formal way of saying $f$ is continuous at $c$? We say that $\lim\limits_{x \to c} f(x) = f(c)$.
I am going to call $f(c)$, $u$. Why? I like that letter (Don't you?). We are also told that $f(c) = u > 0$, in short: it is positive. Now, when we say that $f$ is continuous at $c$ (that limit business), what we mean is that if $x$ is near $c$, $f(x)$ is near $u$. So surely if we pick some $x$ close enough to $c$, we ought to have $f(x)$ close enough to $u$ that it is still positive (even if it's just a little bit less than $u$, values larger than $u$ will be, of course, even more positive).
How close is "close enough"? Here the formal definition of a limit comes to our rescue. When we say $\lim\limits_{x \to c} f(x) = f(c) = u$, what we mean (formally) is given any positive number (even an itty bitty one), which we call $\epsilon$ (apparently, Greek letters are way cooler than English ones), we can find some positive number $\delta$ so that whenever $|x - c| &lt \delta$, then $|f(x) - u| &lt \epsilon$. Since we already know $f$ is continuous, we get "$\delta$" for free, anytime we pick an "$\epsilon$". So in this case, "close enough" means "within $\delta$".
Now, all we need to do, is choose $\epsilon$. Following the steps of the great set-theorist Indiana Jones, we shall "choose wisely". Our choice? We shall use $\epsilon = \frac u 2$. Our reasons are we can rest assured that: 


*

*$u/2$ is just as positive as $u$ is, and 

*It is smaller than $u$. Both of these will be important.
Now from the limit definition, and the continuity of $f$, we get our fat, grubby hands on a $\delta$ that guarantees that as long as $|x - c| &lt \delta$, we can be confident that $|f(x) - u| &lt \frac u 2$. However, what we actually want to do is prove $f$ is positive on such a range of $x$'s. So the absolute value signs aren't fair: that automatically positifies everything. So we take it away: recall that if $t &lt 0 &lt s$, then $|t| = -t > 0 > -s$. This lets us re-write the inequality $|f(x) - u| &lt \frac u 2$ as the double inequality:
$$-\frac u 2 &lt f(x) - u &lt \frac u 2$$
where what we have is honest-to-midwestern-goodness real numbers, no more hiding behind those absolute value signs, no sir. Now we can add $u$ to all three terms, without fear of messing up the inequalities:
$$-\frac u 2 + u &lt f(x) - u + u &lt \frac u 2 + u \iff\frac u 2 &lt f(x) &lt 3\frac u 2$$
The inequality on the right-hand isn't very interesting to us, knowing that $f(x) &lt 3u/2$ doesn't reveal anything about whether or not $f(x)$ is positive. But the one on the left is pure gold: it tells us that on the entire real interval $(x-\delta,x+\delta)$, that $u/2 &lt f(x)$. And $u/2$ is positive, so $f(x)$ is even "more positive". And on that happy note, I bid you adieu.
A: From the definition of continuity you have that $f$ is continuous at $x$ if for every $\varepsilon>0$ there exists $\delta>0$ such that for every $y \in (x-\delta,x+\delta)$ you have $|f(x)-f(y)|&lt\varepsilon$.
This means that you can choose a small neighborhood of $x$ such that $f$ is as close as you want to $f(x)$ on that neighborhood.
In your case, you know that $f(x)>0$. Then you can pick a neighborhood such that the distance from $f(x)$ to $f(y)$ is smaller than $f(x)/2$. This prevents $f(y)$ to be negative on that neighborhood.
Of course you can write all this down using epsilon and deltas, and using the triangle inequality. You should try that.
A: I've already given one Function Monkey-related answer today, so why not another?
Suppose we say $f(x_0) = y_0$; that is, when graph-plotting Function Monkey $f$ happens upon the coconut at $x$-value exactly $x_0$, he throws that coconut to height exactly $y_0$.
Now, in general, the Function Monkey's arm may get quite a workout, throwing coconuts high and low and all over. "Continuity of $f$" (at $x=x_0$) is the property that, when the Function Monkey is "sufficiently close" to $x_0$, he'll throw coconuts that land "arbitrarily close" to height $y_0$. Whatever that means.
Here's what that means: 
When (and only when, so "whenn"?) the function is continuous ... If you give me an allowable (in)accuracy in coconut heights --if you will, an acceptable "$\epsilon$"rror (no matter how small) in computing $y_0$-- then I can construct a leash for the Function Monkey that keeps him from wandering too far away --restricting his "$\delta$"istance from $x_0$-- in such a way that every coconut the leashed Function Monkey can reach will be flung with the desired accuracy. Your flexibility to demand any level of accuracy (apart from dead-on bull's-eyes every time, because that's just too much to ask from a Monkey), and my ability to meet every such demand with an appropriately-sized leash (that allows at least some movement, because pinning a Monkey to a single point would be cruel), means that, whatever wild behavior the Function Monkey may exhibit "globally", his behavior "locally" is quite tame: the graph must "pass through" $(x_0,y_0)$ in a nice --continuous-- way.
To reiterate: For every level of accuracy you choose, I can find a satisfactory leash-length to guarantee that accuracy from the Function Monkey. In the parlance, "For every $\epsilon$, there is a $\delta$!"
In the formal definition of continuity, we specify that $\epsilon > 0$ (to allow some wiggle room $y$-wise). We also specify $\delta > 0$; I say that's because pinning the Function Monkey down is cruel, but the real reason (at least, a reason with a much smaller imaginary component) is that it prevents the smart-aleck choice of $\delta=0$ all the time. (Of course a Function Monkey standing right at $x_0$ will throw his coconut "within any $\epsilon$" of $y_0$ ... because the $x_0$ coconut lands on $y_0$! DUH! That's not helpful!) Since we're interested in showing there are coconuts "near enough" to $x_0$ that land "near" $y_0$, we have to let the Function Monkey wander just a little. (Oh, by the way: the definition's stuff about $|f(x)-y_0|<\epsilon$ and $|x-x_0|<\delta$ is just the formal way of saying "$f(x)$ is within $\epsilon$ of $y_0$" and "$x$ is within $\delta$ of $x_0$". But you knew that.)
Be that as it may ... As others have addressed, your specific problem is one in which the "$c$"oconut has been lobbed some height strictly above the $x$-axis. The word "strictly" is important here, because it means there's some (perhaps small) vertical distance between the "$c$"oconut and the ground; its also means that other coconuts can be not-quite-so-high yet still be above ground. Specifically, if the "$c$"oconut is at height $h$, then any coconut "within $h$" (vertically) of the "$c$"oconut will be above ground. This exactly describes an allowable "$\epsilon$"rror in height ---namely, $\epsilon = h$ (or any positive proper fraction thereof)--- that nevertheless guarantees above-ground-ness; consequently, supposing a continuous function, there must exist some allowable "$\delta$"istance the Function Monkey can wander that guarantees coconut-lobbing $\epsilon$-close to the "$c$"oconut, and that, in turn, guarantees above-ground-ness. As desired.
