Finding the limit by factoring the denominator and canceling

Question

I have the problem $$\lim_{x\to10} \frac{x-3}{x^2+7x-30}$$

If I factor it to $\dfrac{x-3}{(x+10)(x-3)}$ then $x-3$ cancels and I'm left with $0$.

I know the real answer is $1/20$, but why is zero wrong?

Answer 1

Your cancellation method needs an adjustment: $$\frac{a}{a\cdot b} = \frac{1}{b}, \quad \textbf{not} \frac{0}{b}$$ To see why, note that $a=a\cdot 1$, so
$$\frac{a}{a\cdot b} = \frac{a\cdot 1}{a\cdot b}= \frac{1}{b}$$

You may also want to experiment with $\dfrac{3}{2\cdot 3}$, for a concrete example.