Homework Help: Limit Question

1. Dec 14, 2011

Miike012

lim x → 0 sin(x)/x = 1

This doesnt mean that f '(0) = 1 if f(x) = sin(x)/x does it?

2. Dec 14, 2011

Staff: Mentor

No, it's just a limit that's not related to a derivative.

3. Dec 14, 2011

Miike012

But derivates and limits are similar correct?

4. Dec 14, 2011

Staff: Mentor

A derivative is a limit, but a limit is not necessarily a derivative.

5. Dec 14, 2011

Dick

That's a difference quotient for f(x)=sin(x). Not for f(x)=sin(x)/x.

6. Dec 15, 2011

Harrisonized

Definition of a derivative:

df/dx = limΔx→0 [f(x+Δx)-f(x)]/Δx

Changing the variables around, we find:

df/dy = limx→0 [f(y+x)-f(y)]/x

If limx→0 f(y+x)-f(y) = sin(x), then your relation holds. Not really sure what function would satisfy this though. Maybe you can tell me. :P

7. Dec 15, 2011

HallsofIvy

It certainly can be used to show that:

$$\frac{dsin}{dx}(0)= \lim_{x\to 0}\frac{sin(x)- sin(0)}{x}= \lim_{x\to 0} \frac{sin(x)}{x}$$.

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