Why Does lim ( f(a + Δx) - f(a) ) / Δx → 0? Geometrical Meaning & More

  • Thread starter Thread starter Miike012
  • Start date Start date
  • Tags Tags
    Continuity
Miike012
Messages
1,009
Reaction score
0
Why does the book say if f(x) is continuous at a then

lim ( f(a + Δx) - f(a) ) / Δx, that Δx will go to zero. What does that mean geometrically?
Δx→0

More importantly, why would Δx not approach zero if f(x) is not continuous at a?

Im guessing it has something to do with the slopes of the secant lines approaching a definite number therefore Δx will approach 0. However say we were given g(x) = 1/(x - a)

Then:
g'(a) = lim ( 1/(Δx) - 1/0 ) / Δx therefore not only is 1/0 not defined but 1/Δx does not have a limit near 0 in
Δx→0

this case so the limit as Δx → 0 does not exist.

Am I on to something or am I way off base?
 
Last edited:
Physics news on Phys.org
You should try to be a bit clearer in your posts as I'm having trouble figuring out exactly what you're saying/asking.

Geometrically, taking Δx→0 in your difference quotient corresponds to 'drawing' the secant line away from the curve until it becomes a tangent line at the desired point.

In regard to your function g(x)=1/(x - a), this function is not defined at a, thus evaluating its derivative at a makes no sense.
 
I will have to find exactly what the book said but it said something that if f(x) is continuous at a then delta x will go to zero.
 
Miike012 said:
Why does the book say if f(x) is continuous at a then

lim ( f(a + Δx) - f(a) ) / Δx, that Δx will go to zero. What does that mean geometrically?
Δx→0

More importantly, why would Δx not approach zero if f(x) is not continuous at a?
...

The book is only addressing the case in which f(x) is continuous at a. It's not saying anything about the case where f(x) is not continuous at a.

Furthermore, \displaystyle \lim_{\Delta x\to0}\frac{f(a+\Delta x)-f(a)}{\Delta x} is a definition for the derivative of f(x) at x=a. This is only defined if f(x) is continuous at a.

This limit may or may not exist, even if f(x) is continuous at a.

In fact it is possible for this limit to exist if f(x) is not continuous at a, if the discontinuity is removable. But in this case this limit would not be the derivative of f.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top