# 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.

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Thanks a bunch guys, you were very helpful! – Juan May 9 '11 at 4:27
No problem. Could you accept one of the answers so that this question does not keep popping up on the first page? – user17762 May 9 '11 at 4:38

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

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

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

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Using L'Hopital's Rule,

$\displaystyle\lim_{x \to 4} \frac{x-4}{5 - \sqrt{x^2 + 9}} = \displaystyle\lim_{x \to 4} -\frac{\sqrt{x^2 + 9}}{x} = -\frac{5}{4}$

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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.

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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 $-{5\over 4}$.

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Assuming a = numerator conjugated and b = denominator conjugated. You can simplify this by doing the following:

$$\frac{numerator}{denominator} \times \frac{a}{b} \times \frac{b}{a}$$

$$\lim_ {x \to 4} \frac{x - 4}{5 - \sqrt{x^2 + 9}} =$$ $$\lim_ {x \to 4} \frac{x - 4}{5 - \sqrt{x^2 + 9}} \times \frac{x + 4}{5 + \sqrt{x^2 + 9}} \times \frac{5 + \sqrt{x^2 + 9}}{x + 4} =$$ $$\lim_ {x \to 4} \frac{x^2 - 16}{25 - (x^2 + 9)} \times \frac{5 + \sqrt{x^2 + 9}}{x + 4} =$$ $$\lim_ {x \to 4} \frac{x^2 - 16}{-(x^2 - 16)} \times \frac{5 + \sqrt{x^2 + 9}}{x + 4} =$$ $$\lim_ {x \to 4} \frac{1}{-1} \times \frac{5 + \sqrt{x^2 + 9}}{x + 4} =$$ $$\lim_ {x \to 4} -1 \times \frac{5 + \sqrt{x^2 + 9}}{x + 4}$$
$$\lim_ {x \to 4} -1 \times \frac{5 + \sqrt{x^2 + 9}}{x + 4} = -1 \times \frac{5 + \sqrt{4^2 + 9}}{4 + 4} = -1 \times \frac{10}{8} = - \frac{5}{4}$$