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I need to find the tangent line at $(-1, f(-1))$ of the following equation: $$f(x) = \dfrac{x+2}{1-x}-3x$$

I tried this: $$ \lim_{h \to 0} \dfrac{f(-1+h) - f(-1)}{h}$$ which is giving me this: $$\lim_{h \to 0} \dfrac{\dfrac{-1+h+2}{1-1+h} -3(-1+h)-\dfrac{-1+2}{1--1} -3(-1)}{h}=\lim_{h \to 0} \dfrac{\dfrac{h+1}{h} -3h - 0.5}{h}$$ But now, I'm stuck, I can't replace $h$ with $0$ and simplify.

Any help would be appreciated.

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There was a sign error when you substituted. It is not $\frac{-1+h+2}{1-1+h}$. The denominator should be $1-(-1+h)=2-h$. I suggest you use more parentheses. – André Nicolas Nov 24 '12 at 1:56
up vote 1 down vote accepted

The slope of the tangent at $(-1,f(-1))$ can be found by differentiating $f(x)$ and plugging in $x=-1$. Based on your work, it looks like want to evaluate the slope using first principles. Hence, slope at $x=-1$ is given by \begin{align} \lim_{h \to 0} \dfrac{f(-1+h)-f(-1)}h & = \lim_{h \to 0} \dfrac{\left(\dfrac{-1+h+2}{1-(-1+h)}-3(-1+h) \right) - \left(\dfrac{-1+2}{1-(-1)}-3(-1) \right)}h\\ & = \lim_{h \to 0} \dfrac{\left(\dfrac{h+1}{2-h}+3 - 3h \right) - \left(\dfrac{1}{2}+3 \right)}h = \lim_{h \to 0} \dfrac{\left(\dfrac{h+1}{2-h} - 3h \right) - \dfrac{1}{2}}h\\ & = \lim_{h \to 0} \dfrac{\overbrace{\left(\dfrac{h+1}{2-h} - \dfrac{1}{2} \right)}^{\text{Take the l.c.m}} - 3h}h\\ & = \lim_{h \to 0} \dfrac{\left(\dfrac{2h+2-2+h}{4-2h} \right)-3h}h = \lim_{h \to 0} \dfrac{\left(\dfrac{3h}{4-2h} \right)-3h}h\\ & = \lim_{h \to 0} \left(\dfrac3{4-2h} - 3 \right) = \dfrac34 -3 = -\dfrac94 \end{align} Hence, the slope of the tangent at $(-1,f(-1))$ is $-9/4$. You also know that the tangent passes through $(-1,f(-1))$. Note that $$f(-1) = \left(\dfrac{-1+2}{1-(-1)}-3(-1) \right) = \dfrac72$$ Hence, the tangent passes through $(-1,7/2)$ and has slope $-9/4$. The equation of the tangent is $$\dfrac{y-7/2}{x+1} = \dfrac{-9}4 \implies y = \dfrac72 - \dfrac94(x+1) = -\dfrac94x + \dfrac54$$ Hence, the equation is $$4y+9x=5$$

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Thanks for the answer! How did you get from that $\frac{\frac{(h+1)}{(2-h)} - 3h - 1/2}{h}$ to that (((2h+2-2+h)/(4-2h))-3h)/h ? I don't see it :\ – Francis Nov 24 '12 at 2:23
@FrancisLacoste I have added the intermediate step $$\lim_{h \to 0} \dfrac{\overbrace{\left(\dfrac{h+1}{2-h} - \dfrac{1}{2} \right)}^{\text{Take the l.c.m}} - 3h}h$$ Hope it is now clear. – user17762 Nov 24 '12 at 2:32
This is very clear, thank you very much, it's really appreciated ! – Francis Nov 24 '12 at 2:34

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