Why Doesn't f(x + Δx) Approach f(x) as Δx Approaches 0 in Limits?

[haribol]

In the attached picture, the equation for the limit I think is:

\lim_{\Delta x \rightarrow 0} \\\ \frac{f(x+ \Delta x) - f(x)}{\Delta x}

When \Delta x approaches 0, why wouldn't the f(x+ \Delta x) approach f(x)? Because as the point Q approaches P, then wouldn't the y value of point Q also approach that of P? Or am I not understanding the concept?

 

[HallsofIvy]

In the attached picture, the equation for the limit I think is:

\lim_{\Delta x \rightarrow 0} \\\ \frac{f(x+ \Delta x) - f(x)}{\Delta x}

When \Delta x approaches 0, why wouldn't the f(x+ \Delta x) approach f(x)? Because as the point Q approaches P, then wouldn't the y value of point Q also approach that of P? Or am I not understanding the concept?

You appear to be misunderstanding the concept. And, considering that the concept is that of a "derivative", one of the keys to calculus, you really want to understand it!

Yes, of course, f(x+ \Delta x) approaches f(x). If it didn't, since the denominator necessarily approaches 0, the limit of the fraction would not exist!

You may be thinking "If the numerator goes to 0, then the limit must always be 0" but that's only true if the limit of the denominator is NOT 0.
If neither numerator nor denominator goes to 0, that's easy: you can take those limits separately and the limit of the fraction is just one over the other.
If numerator goes to 0 and denominator does not, that's easy: the limit of the fraction is 0.
If numerator does not go to 0 and denominator does, that's easy: the limit of the fraction does not exist.
The only "hard" case, and the interesting one, is exactly what is given here: both numerator and denominator go to 0 and we cannot look at the numerator and denominator separately.

 

[haribol]

I don't quite understand you. According to that graph, then the limit is 0/0?

 

[humanino]

Parameterize \Delta x = x_0 \times p as an evolution with parameter p\rightarrow 0. Then if the numerator goes like f(x+\Delta x) - f(x) \approx s_0 \times p you see that the ratio will tend to \frac{s_0}{x_0} which is non-vanishing, although both the numerator and the denominator go to zero.

 

[HallsofIvy]

No, the limit is not 0/0- that is an "indeterminant" and it doesn't mean anything. What it means is that if you have a fraction such that both numerator and denominator go to 0, then you have to look more closely.

What is the limit of x²/x as x goes to 0? What is the limit of 3x/x as x goes to 0? Think hard about those limits before you start on derivatives.