pdf is defined as $f_X(x)=P(X=x)$ but isn't $P(X=x)=0$ When we define a probability distribution function, we say:
$f_X(x)=P(X=x)$ and thats equal to some function such as a gaussian
But isn't $P(X=x)=0$ for a continuous random variable $X$. 
Is it correct that the height of the pdf function at a specific x represents the likelihood of this $x$.
 A: No. Loosely speaking,
$$f_X(x) = \frac{P(X\in dx)}{dx}.$$
The area around a particular value $x$
is 
$$P(X\in dx) = f_X(x)\,dx.$$
It is the case that $P(X =x) = 0$ wherever the distribution has density.
A: Putting probablyme's answer in words, similar to yours:
The height of the pdf function at a specific $x$ is proportional to the probability of being in the neighbourhood of that $x$.
The main difference being "in the neighbourhood of $x$", instead of "at $x$". Also "proportional", not "the value"
A remark: in English "likelihood" and "probability" are used as loose synonyms; but in mathematics, and in this context of using a function to predict an outcome, you better use "probability". "Likelihood" is mostly used in the reverse case, when some outcome(s) are already known.
A: No. $~$ The probability mass at any point in the support of any real-valued continuous random variable is zero, but it has a probability density there.   The probability density does not represent the "likelyhood" of that point occurring, but rather the gradient, or rate of change, of the Cumulative Distribution Function (CDF) at that point.
$$\begin{align}f_X(x) = & ~ {F}'_X(x) \\[1.5ex] = & ~ \dfrac{\mathrm d \mathsf P(X\leq x)}{\mathrm d x\qquad}\end{align}$$
Then for any $a< b$ : $$\begin{align}\mathsf P(a < X \leq b) = & ~ \int_a^b f_X(x)\operatorname d x\\[1.5ex]  = & ~ {F}_X(b)-{F}_X(a)\end{align}$$
A: What you are confusing is discrete and continuous case. We do not define densities in that way when we are talking about continuous random variables. Actually, we say that $X$ is continuous random variable if there is $f$ such that $$P(a\leq X\leq b) = \int_a^b f(x)\, dx$$ Directly from that definition we do get your claim that $P(X=a) = \int_a^a f(x)\, dx = 0$ for continuous random variables.
So, what exactly is $f(x) = P(X=x)$? This is called probability mass function and makes sense only for discrete random variables, as you notice yourself. This actually can be thought of as probability density function, but not with respect to Lebesgue measure, as in case of continuous random variables, but with respect to counting measure when we have $$P(a\leq X \leq b) = \sum_{x_n
\in [a,b]} P(X=x_n) = \int_a^b f(x)\, d\mu(x)$$
