# Not equivalent at a?

1. Mar 7, 2010

### Shaybay92

When using direct substitution to calculate the limit at x = a some functions are simplified so that x = a is actually defined. For example:

Lim x->1 [(x^2 - 1)/(x-1)]

Limx->1 [(x-1)(x+1)/(x-1)]

Lim x->1 [(x+1)] = 2 (when x=1 is substituted in)

I understand that they can have the same limits despite not both being defined at x=1, however, what I don't get is why the original f(x) isnt defined, but the second one is. How can two equivalent functions not be defined at the same points? Can't all functions be simplified by factorising without jeapordizing where they are actually defined? This makes no sense to me...

2. Mar 7, 2010

### Tinyboss

When you made that simplification, you assumed that $$\frac{x-1}{x-1}=1$$. That's only true when $$x\ne1$$. When x=1, it's not a valid operation, which is why the behavior of the function is changed there.

3. Mar 8, 2010

### HallsofIvy

$$\frac{x^2- 1}{x- 1}$$ is not defined at x= 1 because $$\frac{0}{0}$$ is not defined.