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.

Solution 1:

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 $$\frac{x-4}{5 - \sqrt{x^2+9}} = - \frac{1}{\quad\frac{\sqrt{x^2+9}-5}{x-4}\quad}.$$

Solution 2:

$$\frac{x-4}{5 - \sqrt{x^2 + 9}} = \frac{x-4}{25 - (x^2 + 9)} \times (5 + \sqrt{x^2 + 9}) = \frac{x-4}{16 - x^2} \times (5 + \sqrt{x^2 + 9}) = \frac{5 + \sqrt{x^2 + 9}}{-(x+4)}$$

Hence, $$\lim_{x \rightarrow 4} \frac{x-4}{5 - \sqrt{x^2 + 9}} = \lim_{x \rightarrow 4} \frac{5 + \sqrt{x^2 + 9}}{-(x+4)} = \frac{5+5}{-8} = - \frac{5}{4}$$

Solution 3:

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} $

Solution 4:

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.