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We've been given the definition of a derivative as:

$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$

We are asked to use this to find the derivative of the function $f(x)=\frac{1}{1-x}$ showing every step.

I can get to here: $$f'(x)=\lim_{h\to0}\frac{1}{h-hx-h^2}-\frac{1}{h-hx}$$

When I try using online equation solvers they just jump straight to the answer and I can't figure out how. Wolfram Alpha's step-by-step solution also doesn't give me any intermediate steps between this and the solution:

$$\frac{1}{(x-1)^2}$$

So, my question is, how do I solve a limit like the one above where every value in the denominator approaches 0?

(I'm guessing I'm just missing some algebraic tricks)

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2 Answers 2

up vote 6 down vote accepted

Hint: Bring the expression you got to a common denominator. There will be pleasant cancellation.

Remark: It would have made life easier if you had kept the $h$ "outside," that is, looked at $$\frac{1}{h}\left(\frac{1}{1-x-h} -\frac{1}{1-x}\right).$$ Then same common denominator hint, somewhat less messy.

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Thanks! While both the answers helped together, I gave it to you because you were 16 seconds faster :P –  Josh Oct 16 '12 at 0:08

$$\dfrac{\dfrac1{(1-x-h)} - \dfrac1{(1-x)}}{h} = \dfrac{\dfrac{(1-x) - (1-x-h)}{(1-x)(1-x-h)}}{h} = \dfrac{h}{h(1-x)(1-x-h)} = \dfrac1{(1-x)(1-x-h)}$$ Now apply the limit.

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1  
Thanks! Took me a while to figure out what was going on, and then I was like, d'oh, of course you can multiply the denominators across both functions to find get the same denominator on both sides. Kids, this is what happens when you stop practicing your maths for 7 years. –  Josh Oct 16 '12 at 0:12

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