# Question regarding use of limits

• Jun 6th 2011, 12:55 PM
scounged
Question regarding use of limits
I've been thinking about proving the product rule, and I'm not sure if I can use limits in this way to do it:

A) $f'(x)=\lim_{h\to \0} \frac{f(x+h)-f(x)}{h}~\Rightarrow~f(x+h)=\lim_{h\to \0} f'(x)h+f(x)$

or should it be like this:

B) $f'(x)=\lim_{h\to \0} \frac{f(x+h)-f(x)}{h}~\Rightarrow~\lim_{h \to \0} f(x+h)=\lim_{h\to \0} f'(x)h+f(x)$

As stated in the beginning of this post I was trying to prove the product rule, and consequently set this up:

$F(x)=f(x)g(x)$
$F'(x)=\lim_{h \to \0}\frac{F(x+h)-F(x)}{h}\Rightarrow F'(x)=\lim_{h \to \0} \frac{f(x+h)g(x+h)-f(x)g(x)}{h}$

A) feels a bit suspicious, but if it works(in the case above) it would make my task very easy, and B) seems a little pointless...

I'm feeling a bit puzzled at the moment, and any clarification would be appreciated.
• Jun 6th 2011, 01:08 PM
Prove It
This is a standard proof of the product rule:

\displaystyle \begin{align*}\frac{d}{dx}\left[u(x)v(x)\right] &= \lim_{h \to 0}\frac{u(x + h)v(x +h) - u(x)v(x)}{h} \\ &= \lim_{h \to 0}\frac{u(x + h)v(x + h) - u(x + h)v(x) + u(x + h)v(x) - u(x)v(x)}{h} \\ &= \lim_{h \to 0}\frac{u(x + h)v(x + h) - u(x+ h)v(x)}{h} + \lim_{h \to 0}\frac{u(x +h)v(x) - u(x)v(x)}{h} \\ &= \lim_{h \to 0}u(x + h)\lim_{h \to 0}\frac{v(x + h) - v(x)}{h} + v(x)\lim_{h \to 0}\frac{u(x + h) - u(x)}{h} \\ &= u(x)\frac{dv(x)}{dx} + v(x)\frac{du(x)}{dx}\end{align*}

Q.E.D.
• Jun 6th 2011, 02:11 PM
HallsofIvy
Quote:

Originally Posted by scounged
I've been thinking about proving the product rule, and I'm not sure if I can use limits in this way to do it:

A) $f'(x)=\lim_{h\to \0} \frac{f(x+h)-f(x)}{h}~\Rightarrow~f(x+h)=\lim_{h\to \0} f'(x)h+f(x)$

This is impossible because the left side has an "h" and the right side does not.

Quote:

or should it be like this:

B) $f'(x)=\lim_{h\to \0} \frac{f(x+h)-f(x)}{h}~\Rightarrow~\lim_{h \to \0} f(x+h)=\lim_{h\to \0} f'(x)h+f(x)$
For any h f'(x) is approximately $\frac{f(x+h)- f(x)}{h}$ so that f'(x)h is approximately f(x+h)- f(x), f(x+h) is approximately f'(x)h+ f(x) so, yes, $\lim_{h\to 0} f(x+h)= \lim_{h\to 0} f'(x)h+ f(x)$ (which only says "f(x)= f(x)"!).

Quote:

As stated in the beginning of this post I was trying to prove the product rule, and consequently set this up:

$F(x)=f(x)g(x)$
$F'(x)=\lim_{h \to \0}\frac{F(x+h)-F(x)}{h}\Rightarrow F'(x)=\lim_{h \to \0} \frac{f(x+h)g(x+h)-f(x)g(x)}{h}$

A) feels a bit suspicious, but if it works(in the case above) it would make my task very easy, and B) seems a little pointless...

I'm feeling a bit puzzled at the moment, and any clarification would be appreciated.
• Jun 6th 2011, 08:01 PM
abhishekkgp
Quote:

Originally Posted by scounged
I've been thinking about proving the product rule, and I'm not sure if I can use limits in this way to do it:

A) $f'(x)=\lim_{h\to \0} \frac{f(x+h)-f(x)}{h}~\Rightarrow~f(x+h)=\lim_{h\to \0} f'(x)h+f(x)$
as hallsofivy points out, the last equality in the above line is wrong.. in physics or engineering such equalities are taken to be true since in most practical situations those are approximately true but strictly in many cases that will give huge error.

or should it be like this:

B) $f'(x)=\lim_{h\to \0} \frac{f(x+h)-f(x)}{h}~\Rightarrow~\lim_{h \to \0} f(x+h)=\lim_{h\to \0} f'(x)h+f(x)$

As stated in the beginning of this post I was trying to prove the product rule, and consequently set this up:

$F(x)=f(x)g(x)$
$F'(x)=\lim_{h \to \0}\frac{F(x+h)-F(x)}{h}\Rightarrow F'(x)=\lim_{h \to \0} \frac{f(x+h)g(x+h)-f(x)g(x)}{h}$

A) feels a bit suspicious, but if it works(in the case above) it would make my task very easy, and B) seems a little pointless...

I'm feeling a bit puzzled at the moment, and any clarification would be appreciated.

...
• Jun 7th 2011, 12:47 PM
scounged
Yes, I suspected that I had made an error. Thanks for the clarification.
• Jun 7th 2011, 12:49 PM
scounged
Quote:

Originally Posted by Prove It
This is a standard proof of the product rule:

\displaystyle \begin{align*}\frac{d}{dx}\left[u(x)v(x)\right] &= \lim_{h \to 0}\frac{u(x + h)v(x +h) - u(x)v(x)}{h} \\ &= \lim_{h \to 0}\frac{u(x + h)v(x + h) - u(x + h)v(x) + u(x + h)v(x) - u(x)v(x)}{h} \\ &= \lim_{h \to 0}\frac{u(x + h)v(x + h) - u(x+ h)v(x)}{h} + \lim_{h \to 0}\frac{u(x +h)v(x) - u(x)v(x)}{h} \\ &= \lim_{h \to 0}u(x + h)\lim_{h \to 0}\frac{v(x + h) - v(x)}{h} + v(x)\lim_{h \to 0}\frac{u(x + h) - u(x)}{h} \\ &= u(x)\frac{dv(x)}{dx} + v(x)\frac{du(x)}{dx}\end{align*}

Q.E.D.

Ah, I've seen this one before and I remember that I found it very easy to follow when I was initially introduced to the product rule. Sadly, when trying to recreate it myself I always seem to forget to insert the -u(x+h)v+u(x+h)v. I think that I will grow out of this weird sort of "partial amnesia" eventually though.
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