Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


I am not sure what to do, I have tried factoring everything and using both conjugates, neither options gives me anything usable.

share|cite|improve this question
up vote 5 down vote accepted

First pull out the obvious factor of $x$ in the denominator to get $$\frac{4-\sqrt x}{x(16-x)}\;.$$ The $x$ in the denominator won’t cause any problems in taking the limit, so focus on the rest: $$\frac{4-\sqrt x}{16-x}\;.$$ Notice that each term in the denominator is the square of the corresponding term in the numerator: $$\frac{4-\sqrt x}{16-x}=\frac{4-\sqrt x}{4^2-(\sqrt{x})^2}\;.$$ That last denominator is the difference of two squares; what do you know about factoring such differences?

share|cite|improve this answer
No I forgot the formula. – user138246 Jan 15 '12 at 1:09
@Jordan: David Mitra gave it at the end of his answer. That’s one that you really need to remember: it comes up a lot. – Brian M. Scott Jan 15 '12 at 1:12
For whatever reason I still am not getting the right answer. For the denominator I get $(4+\sqrt{x})(4+\sqrt{x})$ I then cancel out the $(4-\sqrt{x})$ and I am left with $(4+\sqrt{x})$ which is 8. – user138246 Jan 15 '12 at 1:45
@Jordan: That’s fine, but don’t the rest of the expression: I set the $x$ aside to focus on the rest, but it’s still there, so after your cancellation the fraction is $\frac1{x(4+\sqrt{x})}$, and the limit is $\frac1{16\cdot 8}=\frac1{128}$. – Brian M. Scott Jan 15 '12 at 3:41

Let's try to get things to look the same: $$ {4-\sqrt x\over 16x-x^2}={4(1-{\sqrt x\over4})\over 16x(1-{x\over 16} ) }= { 1-{\sqrt x\over4} \over 4x(1-{x\over 16} ) }. $$ Observe that $({\sqrt x\over 4})^2={x\over 16}$ for $x>0$.

Can you see how to take advantage of the formula $a^2-b^2=(a+b)(a-b)$?

share|cite|improve this answer
It’s a perfectly reasonable answer, but I think that introducing the fractions makes it a little uglier than necessary. – Brian M. Scott Jan 15 '12 at 0:44

Alternative Method: (Assuming one knows differentiation)

$$\lim_{x\to16}\dfrac{4-\sqrt{x}}{16x-x^2}=\lim_{x\to16}\frac1x\dfrac{4-\sqrt{x}}{16-x}=\lim_{x\to16}\frac1x\dfrac{\sqrt{16}-\sqrt{x}}{16-x}.$$ By definition, $\lim\limits_{x\to16}\frac{4-\sqrt{x}}{16-x}$ is the derivative of $\sqrt{x}$ evaluated at $x=16$. And since the derivative of $\sqrt{x}$ is $\frac{1}{2\sqrt{x}}$, it follows that its value is: $\frac{1}{2\sqrt{16}}=\frac{1}{8}$. Thus $\lim\limits_{x\to16}\frac{4-\sqrt{x}}{16-x}=\tfrac18$, hence the value of the limit is $\tfrac1{16}\cdot\tfrac18=\tfrac{1}{128}$.

share|cite|improve this answer

Try using L'Hôpital's rule in the case of an indeterminate form.

share|cite|improve this answer
It works, but it’s overkill, and Jordan may not yet have reached l’Hospital’s rule in his course. – Brian M. Scott Jan 15 '12 at 0:45

One way is by a rationalizing substitution: $$ \begin{align} u & = \sqrt{x} \\ \\ u^2 & = x \\ \\ \text{As }x\to16, & u \to 4. \end{align} $$ So $$\lim_{x\to16}\frac{4-\sqrt{x}}{16x-x^2} = \lim_{u\to4}\frac{4-u}{16u^2-u^4} = \lim_{u\to 4} \frac{4-u}{u^2(4-u)(4+u)} = \lim_{u\to4} \frac{1}{u^2(4+u)}=\frac{1}{4^2(4+4)}=\frac{1}{128}.$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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