1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks a bunch guys, you were very helpful! $\endgroup$ – Juan May 9 '11 at 4:27
  • 1
    $\begingroup$ No problem. Could you accept one of the answers so that this question does not keep popping up on the first page? $\endgroup$ – user17762 May 9 '11 at 4:38
5
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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

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

$\endgroup$
4
$\begingroup$

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} $

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $-{5\over 4}$.

$\endgroup$
0
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.