<s>[INST] Why does simplifying a function give it another limit I'm asked: 
$$\lim_{x\to 1} \frac{x^3 - 1}{x^2 + 2x -3}$$
This does obviously not evaluate since the denominator equals $0$. The solution is to: 
$$\lim_{x\to 1} \frac{(x-1)(x^2+x+1)}{(x-1)(x+3)}$$
$$\lim_{x\to 1} \frac{x^2 + x + 1}{x + 3}$$
$$\frac{1+1+1}{1+3} = \frac{3}{4}$$
My question: what is actually happening? How can simplifying a function give it another limit? Is it a complete other function and if so why would it be relevant to our original question? 
 [/INST] It's not that the simplified function has a different limit, it's that the limit of the original expression cannot be found by evaluation. In other words, write
$$
f(x) = \frac{x^{3} - 1}{x^{2} + 2x - 3},\qquad
g(x) = \frac{x^{2} + x + 1}{x + 3}
  = \begin{cases}
  f(x) & x \neq 1, \\
  \frac{3}{4} & x = 1.
\end{cases}
$$
Since $g$ is a quotient of polynomials with denominator non-zero at $x = 1$, its limit at $1$ can be found by evaluation. The limit of $f$ at $1$ has the same value, since $f(x) = g(x)$ for all $x \neq 1$.
</s>