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 trying to study for a test I have tomorrow but I can't even interpret what is happening in my book.

Find an equation of the tangent line at the hyperbola $y=\dfrac3{x}$ at the point $(3,1)$

Let $f(x) = \dfrac3{x}$ then the slope of the tangent at $(3,1)$ is:

$$m = \lim _{h \to 0} \frac{f(3+h)-f(3)}{h} = \lim _{h \to 0} \frac{\frac{3}{3+h} - 1}{h} = \lim _{h \to 0} \frac{\frac{3-(3+h)}{3+h}}{h} = \lim _{h \to 0} -\frac{h}{h(3+h)} = \lim _{h \to 0} \frac{-1}{3+h} = -\frac{1}{3}$$

I do not follow what happened from the first to second step, they somehow manipulate the problem to make the number into the denominator over 3. Not sure how to that but it does need seem like good algebra especially considering h was not in any way manipulated which should change the problem.

share|cite|improve this question
$\dfrac{\frac3{3+h}-1}{h}=\dfrac{\frac{3-3-h}{3+h}}{h}=\frac{-h}{h(3+h)}$... can you see what cancels? – J. M. Sep 11 '11 at 16:33
I don't even see what happened between the two problems. It doesn't seem possible with proper algebra. – user138246 Sep 11 '11 at 16:38
Which two problems? – J. M. Sep 11 '11 at 16:39
Never mind I got it. – user138246 Sep 11 '11 at 16:40
The $-1$ became $-\frac{3+h}{3+h}$, then the preceding $3$ was put over the same denominator. – Ross Millikan Sep 11 '11 at 16:41
up vote 2 down vote accepted

From the first to second step we used the definition of $f$ to evaluate $f(3+h)$ and $f(3)$. From the second to the third we used $-1=-\frac{3+h}{3+h}$ then combined the two fractions in the numerator over the common denominator. From the third to the fourth the $3$ and $-3$ are cancelled and numerator and denominator are multiplied by $\frac{1}{h}$. Then the $h$'s are cancelled (as $h \ne 0$) and the limit is taken.

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.