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I don't understand quiet good something here, for example if we want to find the derivative of the function $\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(h)}{h} $ and if we compute it from the function: $ f(x) = 12 + 7x $ We get that the derivative of $f(x)$ is equal to $$\lim_{h \to 0} \frac{7h}{h}$$ But I thought that we can't divide by zero (here we cancel 0 over 0), I'm I wrong or $\displaystyle \frac{0}{0}$ equals 1?

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Small correction: The definition is either $\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ or $\lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0}$. –  Javier Badia Sep 7 '13 at 21:17
    
Thank you, I will correct it. –  Rihanna Sep 7 '13 at 21:20
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5 Answers

up vote 3 down vote accepted

You are correct in saying that we can't divide by zero. But since we're taking a limit, we aren't dividing by zero.

Remember: a limit only cares what happens near the point, not at the point. So if $h=0$, then $\frac{7h}{h}$ would be undefined. But "$h$ goes to $0$" means that it is close to $0$ but not equal to it. Since for every $h\neq0$, $\frac{7h}{h} = 7$.

The basic idea here is that we don't want to evaluate $\frac{7h}{h}$ at $0$. We want to approach it. And as $h$ becomes closer and closer to $0$, $\frac{7h}{h}$ becomes closer and closer to $7$ (in fact, it is constantly $7$), and so we say that $\lim_{h \to 0}\frac{7h}{h} = 7$, since we don't care what happens at $h=0$, only near it.

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Thank you! This is so clear! What AN ANSWER! Thank you so much! –  Rihanna Sep 7 '13 at 21:25
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The whole point of the limit operation is that it avoids any bad behaviour of a function around the given point. We don't care what the function value is, nor whether it's even defined at a given point.

In your case, so long as $h \ne 0$, we can cancel to find that $\frac {7h}{h} = 7$; it doesn't matter that $\frac{7h}{h}$ isn't defined at $0$.

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But then it violates what I taught in Algebra; –  Rihanna Sep 7 '13 at 21:19
    
@Rihanna How so? The limit doesn't care what happens when $h = 0$, and there is no division by zero. –  T. Bongers Sep 7 '13 at 21:20
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The limit is defined as:

We say $\lim_{x \rightarrow c} f(x)=L$ if for all $\varepsilon>0$ there exists a $\delta>0$ such that for all $x$, if $0<|x-c|<\delta$ then $0<|f(x)-L|<\varepsilon$.

In this case, we have $c=0$, so when $x=0$ the condition $$0<|x-c|$$ is not satisfied.


Note that $$\lim_{h \rightarrow 0} \frac{7h}{h} \neq \frac{\lim_{h \rightarrow 0} 7h}{\lim_{h \rightarrow 0} h}$$ where there would be the $\frac{0}{0}$ problem you mention since we have $$\lim_{h \rightarrow 0} 7h=0=\lim_{h \rightarrow 0} h.$$

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If you're familiar with the $\varepsilon-\delta$ definition of a limit, then you can prove that if two functions $f$ and $g$ that satisfy $f(x) = g(x)$ for all $x \neq c$ in an open interval containing $c$, and $\lim_{x \to c}g(x)$ exists, then $$\lim_{x \to c}g(x) = \lim_{x \to c}f(x)$$

To prove this (feel free to skip this part if you don't care), call $\lim_{x \to c} g(x) = L$, and then by definition for each $\varepsilon > 0$ there exists a $\delta > 0$ such that $f(x) = g(x)$ in the intervals $(c - \delta, c)$ and $(c, c+\delta)$, and $$0 < |x - c| < \delta \implies |g(x) - L| < \varepsilon$$

But then because $f(x) = g(x)$ for all $x$ in the interval other than $x = c$, you know that $$0 < |x - c| < \delta \implies |f(x) - L| < \varepsilon$$ so you can say that $$\lim_{x \to c}f(x) = L$$


This relates to your question because the functions $f(x) = \frac{7h}{h}$ and $g(x) = 7$ only differ at $h = 0$, which combined with the theorem above shows that: $$\lim_{h \to 0}7=7 \implies \lim_{h \to 0} \frac{7h}{h} = 7$$

In general this theorem is what justifies being able to cancel off variables in a limit.

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The limit $\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$ always involves $0/0$. You can't divide $0$ by $0$, but that does not mean the limit does not exist.

Derivatives are about instantaneous rates of change. If a car goes $60$ miles in $2$ hours, its average speed during that time is $\dfrac{60\cdot\text{mile}}{2\cdot\text{hour}}$. It speed at a particular instant might therefore appear to be $\dfrac{0\cdot\text{mile}}{0\cdot\text{hour}}$. Derivatives tell you the speed at an instant, so they always involve $0/0$ in that way.

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