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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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    No problem. Could you accept one of the answers so that this question does not keep popping up on the first page?2011-05-09

6 Answers 6

5

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

4

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

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

4

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} $

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.

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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 $-{5\over 4}$.

0

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} $