Is the value of a probability density function for a given input a point, a range, or both? This post says

A PDF is used to specify the probability of the random variable falling within a particular range of values, as opposed to taking on any one value. 

Is it true? 
this is the PDF of the standard normal distribution.
$$\varphi(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$$
plug in x=0 into the formula above, I can get the probability of taking on one value.
Does that post mean the PDF could be used both for point and  interval?
 A: Elaborating a bit on Trisoloriansunscreen's answer: it's very much true that you only got a probability density function. I'd like to draw an analogy for you. Imagine you have a 3D object, say some complex spaceship, and you know the mass density at every point.
For example, some parts of the spaceship might contain water, which has a mass density of $997 \frac{\text{g}}{\text{l}}$. Does this already tell you anything about the mass of the whole spaceship? No, it does not! Precisely because you only know this value at a specific point. You've got no information on how much water there actually is. It might be $1\ \text{ml}$ or $1\ \text{l}$.
Now suppose you know the amount of water, let's say $2\ \text{l}$. By simple multiplication $997 \frac{\text{g}}{\text{l}} \cdot 2\ \text{l}$, you get roughly $1994\ \text{g}$. I would like to make the point that you just did integration in disguise! Consider the following picture:

The mass you calculated is just the greenly shaded rectangular area. This was only doable as a simple multiplication because the mass density was constant for the amount of water considered and thus yielded a rectangular area.
What if you had mixed forms of water, e.g. some gaseous, some liquid, some in varying temperatures and so on? It could look like this:

Now for computing the mass you would need to integrate that mass density function over the amount of water. Do you see the parallel to probability density functions now? To get an actual probability (cf. mass) you need to integrate the probability density (cf. mass density) over some domain.
A: The citation is true. When you plug $x=0$ to the PDF function, you do NOT get the probability of taking this particular  value. The resulting number is probability density which is not a probability. The probability of taking exactly $x=0$ is zero (consider the infinite number of similarly-likely values in the tiny interval $x\in[0,10^{-100}]$).  
To further convince yourself that this $\varphi(x)$ cannot be a probability, consider decreasing the standard deviation of your normal distribution from $\sigma = 1$ to $\sigma = \frac{1}{100}$. Now, $\varphi(0)=\frac{100}{\sqrt{2\pi}}$ - much more than one. Not a probability.
