Hi ryu1!

You gave a very good example with f(x) = x-1, and g(x) = 2x-2.

However, your limit would be:

$\displaystyle \lim_{x \to 1} \frac {f(x)}{g(x)} = \lim_{x \to 1} \frac {x-1}{2x-2} = \lim_{x \to 1} \frac {(x-1)}{2(x-1)}$

Now your teacher was also right.

You can algebraically manipulate this to make the 0 disappear from the denominator.

Just divide both numerator and denominator by (x-1).

This is allowed for any $\displaystyle x \ne 1$ and this is what a limit is for: to predict where the value is going to from its context, even if it is not defined at 1.

The result is:

$\displaystyle \lim_{x \to 1} \frac {f(x)}{g(x)} = \lim_{x \to 1} \frac {(x-1)}{2(x-1)}= \lim_{x \to 1} \frac {1}{2} = \frac 1 2$

In particular this

*disproves* the statement $\displaystyle \lim_{x \to 1} \frac {f(x)}{g(x)}=1$.