A question about the derivative

[Flumpster]

Homework Statement

Generally the derivative has the limit x-- h applied to the whole thing like
$$\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$

I'm guessing you can't express it as

$$\frac{\lim_{h\to 0} f(x+h)-f(x)}{\lim_{h\to 0} h}$$

because the quotient rule for limits doesn't hold when the limit of bottom part of the fraction equals 0.

Can you express it like this though?

$$\frac{f(x +\lim_{h\to 0} h)-f(x)}{\lim_{h\to 0} h}$$

In other words, does $$\lim_{h\to 0} \frac{f(x+h)-f(x)}{h} = \frac{f(x +\lim_{h\to 0} h)-f(x)}{\lim_{h\to 0} h}$$ ?

 

[SammyS]

Homework Statement

Generally the derivative has the limit x-- h applied to the whole thing like
$$\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$$

I'm guessing you can't express it as

$$\frac{\lim_{h\to 0} f(x+h)-f(x)}{\lim_{h\to 0} h}$$

because the quotient rule for limits doesn't hold when the limit of bottom part of the fraction equals 0.

Can you express it like this though?

$$\frac{f(x +\lim_{h\to 0} h)-f(x)}{\lim_{h\to 0} h}$$

In other words, does $$\lim_{h\to 0} \frac{f(x+h)-f(x)}{h} = \frac{f(x +\lim_{h\to 0} h)-f(x)}{\lim_{h\to 0} h}$$ ?

The limit of the denominator is zero in that expression too. So, no, you can't do that.

 

[Mark44]

Besides what SammyS said, you can't in general "distribute" the limit operation into a function.

IOW, it's generally not true that
$\lim_{h \to 0} f(x + h) = f(x + \lim_{h \to 0} h)$

 

[Bennett.F.L]

Your latter formula means the denominator and the numerator are not synchronized when h--->0
Could you understand me?

 

[Mark44]

Your latter formula means the denominator and the numerator are not synchronized when h--->0
Could you understand me?

Who is this directed to, and what do you mean?

 

[Flumpster]

Thanks, Mark44 and SammyS :)
Bennett.F.L, I'm sorry, I'm not sure what you mean. It's ok though, I think I got it.