Limit of sin(x)/x as x Approaches 0: Does it Equal 1?

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In summary, the conversation discusses the relationship between limits and derivatives, specifically for the function f(x) = sin(x)/x. While a limit can be used to find a derivative, not all limits necessarily represent derivatives. The definition of a derivative is also mentioned, which involves a limit. Ultimately, it is stated that the limit of sin(x)/x does not necessarily equal 1, but it can be used to show that the derivative of f(x) = sin(x)/x at x = 0 is equal to 1.
  • #1
Miike012
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lim x → 0 sin(x)/x = 1

This doesn't mean that f '(0) = 1 if f(x) = sin(x)/x does it?
 
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  • #2
Miike012 said:
lim x → 0 sin(x)/x = 1

This doesn't mean that f '(0) = 1 if f(x) = sin(x)/x does it?
No, it's just a limit that's not related to a derivative.
 
  • #3
But derivates and limits are similar correct?
 
  • #4
A derivative is a limit, but a limit is not necessarily a derivative.
 
  • #5
Miike012 said:
lim x → 0 sin(x)/x = 1

This doesn't mean that f '(0) = 1 if f(x) = sin(x)/x does it?

That's a difference quotient for f(x)=sin(x). Not for f(x)=sin(x)/x.
 
  • #6
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
Miike012 said:
lim x → 0 sin(x)/x = 1

This doesn't mean that f '(0) = 1 if f(x) = sin(x)/x does it?
It certainly can be used to show that:

[tex]\frac{dsin}{dx}(0)= \lim_{x\to 0}\frac{sin(x)- sin(0)}{x}= \lim_{x\to 0} \frac{sin(x)}{x}[/tex].
 

FAQ: Limit of sin(x)/x as x Approaches 0: Does it Equal 1?

What is the limit of sin(x)/x as x approaches 0?

The limit of sin(x)/x as x approaches 0 is equal to 1.

Why is the limit of sin(x)/x as x approaches 0 equal to 1?

This limit is equal to 1 because as x approaches 0, the ratio sin(x)/x approaches the value of 1. This can be proven using the squeeze theorem and the fact that sin(x) is always between -1 and 1.

How can we prove that the limit of sin(x)/x as x approaches 0 is equal to 1?

One way to prove this limit is to use the squeeze theorem. By setting up two other limits, one for cos(x)/x and another for 1/x, and using the fact that sin(x) is between -1 and 1, we can show that the limit of sin(x)/x must also be equal to 1.

Is the limit of sin(x)/x as x approaches 0 always equal to 1?

Yes, the limit of sin(x)/x as x approaches 0 is always equal to 1. This is a fundamental limit in calculus and is used in many other mathematical proofs and applications.

Can the limit of sin(x)/x as x approaches 0 be solved using L'Hopital's rule?

No, L'Hopital's rule cannot be used to solve this limit. L'Hopital's rule only applies to limits of the form 0/0 or infinity/infinity, but in this limit, we have a ratio of two functions that both approach 0 as x approaches 0.

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