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.
 A: Alternatively, if you know derivatives, notice that
$$\lim_{x\to 4}\frac{\sqrt{x^2+9}-5}{x-4}$$
is, by definition, $f'(4)$ with  $f(x) = \sqrt{x^2+9}$. Since 
$$f'(x) = \frac{x}{\sqrt{x^2+9}}$$
then $f'(4) = \frac{4}{5}$. The limit you want is the negative reciprocal, since
$$\frac{x-4}{5 - \sqrt{x^2+9}} = - \frac{1}{\quad\frac{\sqrt{x^2+9}-5}{x-4}\quad}.$$
A: $$\frac{x-4}{5 - \sqrt{x^2 + 9}} = \frac{x-4}{25 - (x^2 + 9)} \times (5 + \sqrt{x^2 + 9}) = \frac{x-4}{16 - x^2} \times (5 + \sqrt{x^2 + 9}) = \frac{5 + \sqrt{x^2 + 9}}{-(x+4)}$$
Hence, $$\lim_{x \rightarrow 4} \frac{x-4}{5 - \sqrt{x^2 + 9}} = \lim_{x \rightarrow 4} \frac{5 + \sqrt{x^2 + 9}}{-(x+4)} = \frac{5+5}{-8} = - \frac{5}{4}$$
A: Using L'Hopital's Rule,
$ \displaystyle\lim_{x \to 4} \frac{x-4}{5 - \sqrt{x^2 + 9}} = \displaystyle\lim_{x \to 4} -\frac{\sqrt{x^2 + 9}}{x} = -\frac{5}{4} $
A: HINT $\rm\displaystyle\quad \frac{x\ -\ a}{\sqrt{f(x)}-\sqrt{f(a)}}\ =\ \frac{\sqrt{f(x)}\:+\sqrt{f(a)}}{\frac{f(x)\ -\: \ f(a)}{x\ -\ a}}\ \to\ \frac{2\ \sqrt{f(a)}}{f{\:'}(a)}\ $ as $\rm\:\ x\to a$
Your problem is simply the negative of the special case $\rm\  f(x) = x^2+9\:,\ \ a = 4\:.$
Alternatively, instead of rationalizing the denominator as above, invert the fraction in order to recognize the limit as a first derivative. For further examples of this technique see my prior posts.
A: For all $x\ne 4$ one has
$${x-4 \over 5-\sqrt{x^2+9}}={(x-4)(5+\sqrt{x^2+9}) \over (5-\sqrt{x^2+9})(5+\sqrt{x^2+9})}={(x-4)(5+\sqrt{x^2+9})\over 16-x^2}=-{5+\sqrt{x^2+9}\over x+4}\ .$$
Here the right side is continuous at $x=4$ and has the value $-{5\over 4}$ there. This means that the limit you want is $-{5\over 4}$.
A: 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
$$
\lim_ {x \to 4} -1 \times \frac{5 + \sqrt{x^2 + 9}}{x + 4} = -1 \times \frac{5 + \sqrt{4^2 + 9}}{4 + 4} = -1 \times \frac{10}{8} = - \frac{5}{4}
$$
