A function $f(x)$ is continuous at $c$ if and only if three things happen:
- The function is defined at $c$.
- The limit of $f(x)$ as $x\to c$ exists.
- The value of the function at $c$ equals the value of the limit: $\lim_{x\to c}f(x) = f(c).$
So just from "the limit of a function as it tends to $c$ is $0$" you cannot tell whether the function is continuous or not: you need the value of the function at $c$ to also be $0$ to be able to conclude that.
In the special case of $f(x) = \cos(x)$ at $x=\frac{\pi}{2}$ (measuring in radians):
- $f(x)$ is defined at $x=\frac{\pi}{2}$.
- The limit of $\cos(x)$ as $x\to\frac{\pi}{2}$ exists: $\lim_{x\to\frac{\pi}{2}} \cos x = 0.$
- The value of the limit agrees with the value of the function, since $\cos(\frac{\pi}{2}) = 0$. Since $\lim_{x\to 0}f(x) = f(0),$ then you can conclude that $f(x)$ is continuous at $0$.
That is: it's not just the numerical value of the limit that matters, it is whether that numerical value agrees with the value of the function at that point that matters.
To the new question: the function $f(x) = \frac{x-36}{\sqrt{x}-36}.$
At $x=36$, the function is defined, with value $0$. The limit of the function can be computed using limit laws, since $\sqrt{x}$ is continuous at $36$. We get: $\begin{align*} \lim_{x\to 36}f(x) &= \lim_{x\to36}\frac{x-36}{\sqrt{x}-36} \\ &= \frac{\lim_{x\to 36}(x-36)}{\lim\limits_{x\to36}(\sqrt{x}-36)}\\ &= \frac{36-36}{\sqrt{36}-36} = \frac{0}{-30} = 0. \end{align*}$ Since $f(x)$ is defined at $36$, has a limit as $x\to 36$, and the value of the limit equals the value of the function, then the function is continuous at $x=36$.
If you meant $f(x) = \dfrac{x-36}{\sqrt{x}-6}$, then this function is not continuous at $x=36$, because it is not defined at $x=36$.