proof of chain rule

**pamparana** · Jun21-09, 04:41 PM

Hello everyone,

I was looking at the proof of chain rule as posted here:

http://web.mit.edu/wwmath/calculus/d...ain-proof.html

I am having trouble understanding why delta(u) tends to 0 as delta(x) tends to 0. Can someone point out to me why that is so?

Many thanks,

Luca

**qntty** · Jun21-09, 04:49 PM

Because the limit $LaTeX Code: \\lim_{\\Delta x \\to 0} \\frac {\\Delta u}{\\Delta x}$ must exist by hypothesis, and the only way that can happen is if $LaTeX Code: \\Delta u$ decreases as $LaTeX Code: \\Delta x$ decreases. The limit doesn't need to be 1 because the rate that the denominator and numerator decrease can differ, but it does need to be finite. Think of what would happen if $LaTeX Code: \\Delta u$ approached a nonzero number or diverged to infinity; the limit would also diverge or not exist.

**slider142** · Jun21-09, 05:21 PM

Originally Posted by pamparana

Hello everyone,

I was looking at the proof of chain rule as posted here:

http://web.mit.edu/wwmath/calculus/d...ain-proof.html

I am having trouble understanding why delta(u) tends to 0 as delta(x) tends to 0. Can someone point out to me why that is so?

Many thanks,

Luca

u is continuous (at at least one point, the point where it is differentiable), which means
$LaTeX Code: \\lim_{x\\rightarrow a} u(x) = u(a)$
for all constants 'a' at which u is continuous which is equivalent to your statement (u(x) approaches u(a) as x approaches a).
Expanding the deltas in your limit we have the statement
$LaTeX Code: \\lim_{x\\rightarrow x_0} \\frac{u(x) - u(x_0)}{x - x_0}$
where a = x₀.