This answer consists of two parts, an informal part that uses a calculator, and a more formal part.
What the calculator can tell us: Set your calculator to radian mode. Let $$f(x)=\frac{\sin x}{5x}.$$ Now use the calculator to try to calculate $f(0)$. The calculator will get very upset. Mine shows an E and refuses to do anything until I reset it. That is because our function $f(x)$ is not defined at $x=0$.
In our limit calculation, we want to find what number (if any) is approached by $f(x)$ as $x$ approaches $0$. So let us evaluate $f(x)$ for various $x$ that are near to $0$, but not equal to $0$.
I will start by finding $f(0.1)$. My calculator says that $f(0.1)$ is approximately $0.1996668$. Your calculator, which is undoubtedly fancier, may give an answer to more decimal places.
Now let's calculate $f(-0.05)$. My calculator says that $f(-0.05)$ is approximately $0.1999167$. Next we calculate say $f(0.003)$. My calculator says that $f(0.003)$ is approximately $0.1999997$.
Next we calculate $f(0.0001)$. My calculator says this is approximately $0.2$. Of course, $f(0.0001)$ is not exactly equal to $0.2$, but to the limit of accuracy of the display, that is the value.
If we look at these calculations, it seems sensible to believe that as $x$ approaches $0$, $f(x)$ approaches $0.2$. That is indeed the case.
But you are undoubtedly expected to give some sort of formal argument, not a plausible conjecture based on a few calculations.
A more formal argument: I am sure that your course notes, and your textbook, discuss in detail the key fact that you are expected to use. This key fact is that $$\lim_{x\to 0}\frac{\sin x}{x}=1.$$
The book, and probably the notes, give a justification of this fact. The details depend on exactly how the sine function is defined. In the majority of calculus courses, the justification is geometric. It uses a diagram, and either a comparison of areas or of lengths. Do look at the details. But from the time the key fact is proved, you can use it freely in further calculations.
Note that $$\frac{\sin x}{5x}=\frac{1}{5}\frac{\sin x}{x}.$$ Let $x$ approach $0$. As $x$ approaches $0$, $\frac{\sin x}{x}$ approaches $1$ (this is our key fact). Thus $$\lim_{x\to 0}\frac{\sin x}{5x}=\lim_{x\to 0}\frac{1}{5}\frac{\sin x}{x}=\frac{1}{5}\cdot 1=\frac{1}{5}.$$
Comment: Very informally, when $x$ is close to but not equal to $0$, $\frac{\sin x}{x}$ is roughly $1$, so $\frac{\sin x}{5x}$ is roughly $\frac{1}{5}$. This level of informality is probably not acceptable in your course. And it shouldn't be, for all too often a very informal argument is simply wrong, and gives the wrong answer. However, it can tell you what's really going on. Then you do a more formal verification, as in the write-up above.