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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.

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$\dfrac{\frac3{3+h}-1}{h}=\dfrac{\frac{3-3-h}{3+h}}{h}=\frac{-h}{h(3+h)}$... 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
2  
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

1 Answer 1

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.

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