<s>[INST] Solving indetermination in limit I'm trying to solve this limit, but I can't get it out of a $\frac{0}{0}$ indetermination:
$$\displaystyle \lim_{x \to 4} \; \frac{x-4}{5-\sqrt{x^2+9}}$$
Maybe there is something I'm missing. Thanks a lot in advanced.
 [/INST] Assuming a = numerator conjugated and b = denominator conjugated. You can simplify this by doing the following:
$$
\frac{numerator}{denominator} \times \frac{a}{b} \times \frac{b}{a}
$$
Applying this in your case
$$
\lim_ {x \to 4} \frac{x - 4}{5 - \sqrt{x^2 + 9}} =
$$
$$
\lim_ {x \to 4} \frac{x - 4}{5 - \sqrt{x^2 + 9}} \times \frac{x + 4}{5 + \sqrt{x^2 + 9}} \times \frac{5 + \sqrt{x^2 + 9}}{x + 4} =
$$
$$
\lim_ {x \to 4} \frac{x^2 - 16}{25 - (x^2 + 9)} \times \frac{5 + \sqrt{x^2 + 9}}{x + 4} = $$
$$
\lim_ {x \to 4} \frac{x^2 - 16}{-(x^2 - 16)} \times \frac{5 + \sqrt{x^2 + 9}}{x + 4} =
$$
$$
\lim_ {x \to 4} \frac{1}{-1} \times \frac{5 + \sqrt{x^2 + 9}}{x + 4} = 
$$
$$
\lim_ {x \to 4} -1 \times \frac{5 + \sqrt{x^2 + 9}}{x + 4}
$$
Now you can apply the Direct Substitution Property
$$
\lim_ {x \to 4} -1 \times \frac{5 + \sqrt{x^2 + 9}}{x + 4} = -1 \times \frac{5 + \sqrt{4^2 + 9}}{4 + 4} = -1 \times \frac{10}{8} = - \frac{5}{4}
$$
</s>