Two limit theorems

This theorem is true by virtue of the earlier limit laws. By applying the product rule, we can get $\displaystyle\lim_{x \to a} x^n = a^n$. Combining this with our rule for multiples and sums gives the theorem for polynomials. Combining that with our rule for quotients gives the theorem for rational functions.

In practice, the theorem says that whenever $f$ is a polynomial or rational function, we can evaluate $f$ at $a$, and if this value exists, it is the limit as $x$ approaches $a$.

For example, if we wish to evaluate $$\lim_{x\rightarrow 3} (x^2-4),$$ we simply plug $3$ into $x^2-4$, getting 5. Another example: $$\lim_{x\rightarrow 4}\frac{x-2}{x+2}=\frac{4-2}{4+2}=\frac{1}{3}.$$ The conclusion:  if you plug $a$ into $f$ and there are no issues (dividing by zero, negative under a square root, etc.) most likely that is the limit value.

When $x$ is close to $a$ (but not equal to $a$), $g(x)=f(x)$, so $g(x)$ is close to whatever $f(x)$ is close to. This seems like a trivial result, but it is very useful for evaluating limits of ratios where both the numerator and denominator go to zero. For instance $(x^2-1)/(x-1) = x+1$ whenever $x \ne 1$, so $$\lim_{x \to 1} \frac{x^2-1}{x-1} = \lim_{x \to 1} (x+1) = 1+1 = 2.$$

The conclusion:  If you can factor the top and bottom and cancel a factor from both, and the factor is zero at $x=a$, then the original function and the function you did more work on are the same except at $x=a$, so their limits are the same as $x\to a$.  To restate slightly differently, notice that this limit was of the form $\tfrac00$, so we needed to do more work.  The work we did was getting rid of common factors in the top and bottom, which we could do because of the second theorem above.