Why does delta(u) tend to 0 as delta(x) tends to 0 in the chain rule?

[pamparana]

Hello everyone,

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

http://web.mit.edu/wwmath/calculus/differentiation/chain-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]

Because the limit $$\lim_{\Delta x \to 0} \frac {\Delta u}{\Delta x}$$ must exist by hypothesis, and the only way that can happen is if $$\Delta u$$ decreases as $$\Delta x$$ 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 $$\Delta u$$ approached a nonzero number or diverged to infinity; the limit would also diverge or not exist.

 

[slider142]

Hello everyone,

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

http://web.mit.edu/wwmath/calculus/differentiation/chain-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
$$\lim_{x\rightarrow a} u(x) = u(a)$$
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
$$\lim_{x\rightarrow x_0} \frac{u(x) - u(x_0)}{x - x_0}$$
where a = x₀.