Canceling zero while evaluating a limit I am extremely sorry if my title doesn't actually matches with the question that i have asked. I was solving a limit. It was as follows:
We are given that $ \lim_{x \to 2} \frac{f(x) - 5}{x-2} = 3$ and we have to find   $\lim_{x \to 2}f(x)$. I did this:
$$\lim_{x \to 2}\frac{f(x) - 5}{(x-2)} = 3$$
$$\implies \frac{\lim_{x \to 2} f(x) -5}{\lim_{x \to 2}(x-2)} = 3$$
$$\implies \frac{\lim_{x \to 2} f(x) -\lim_{x \to 2}5}{\lim_{x \to 2}(x-2)} = 3$$
$$\implies \frac{\lim_{x \to 2} f(x) -\lim_{x \to 2}5}{\lim_{x \to 2}(x-2)}*\lim_{x \to 2}(x-2) = 3*\lim_{x \to 2}(x-2) $$
$$\implies \lim_{x \to 2} f(x) -\lim_{x \to 2}5 = 3*\lim_{x \to 2}(x-2)$$
$$\implies \lim_{x \to 2} f(x) -5 = 3*0$$
$$\implies \lim_{x \to 2} f(x) -5 = 0$$
$$\implies \lim_{x \to 2} f(x) = 5$$
and the answer matched with the answer given. But in step $4$, I canceled $ \lim_{x \to 2}(x-2)$ which actually zero. So I don't think that I can do this. Still I found the answer. Is my method right then how is it? and if wrong , is there any other method to find this limit?
 A: In order to give a valid justification, you must correctly apply the theorems on limits. To start with, here are two limits that you know are true: the first is known by hypothesis; and the second is probably already well known to you:
$$\lim_{x \to 2} \frac{f(x)-5}{x-2} = 3 \qquad\lim_{x \to 2} (x-2) = 0
$$
Now apply the following theorem on limit of a product:

Theorem: If $\lim_{x \to a} g(x)$ and $\lim_{x \to a} h(x)$ both exist then $\lim_{x \to a} (g(x) \cdot h(x))$ exists and
$$\lim_{x \to a} (g(x) \cdot h(x)) = \bigl( \lim_{x \to a} g(x) \bigr) \cdot \bigl(\lim_{x \to a} h(x) \bigr)
$$

Applying this theorem using $g(x) = \frac{f(x)-5}{x-2}$ and $h(x)=x-2$, whose limits as $x \to 2$ are known to exist, you may conclude that
\begin{align*}
\lim_{x \to 2} (f(x)-5) &= \lim_{x \to 2} \left( \frac{f(x)-5}{x-2} \cdot (x-2) \right) \\
&= \lim_{x \to 2} \frac{f(x)-5}{x-2}  \cdot \lim_{x \to 2} (x-2) \\
&= 3 \cdot 0 \\
&= 0
\end{align*}
Then, knowing that $\lim_{x \to 2} (f(x)-5)=0$, it is a small step to conclude that $\lim_{x \to 2} f(x)=5$ (you can fully justify that step by applying the theorem on limit of a sum).
