Is the Limit of Sin x/x=1 Proven in Elementary Calculus?

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Discussion Overview

The discussion revolves around the limit of the function sin(x)/x as x approaches 0, specifically whether it is equivalent to the limit of sin(x) equating to x. Participants explore the implications of these limits and the conditions under which they hold true.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant states that the limit (lim x-->0) (sin x)/x = 1 is known from elementary calculus and questions if it is equivalent to (lim x-->0) sin x = x.
  • Another participant argues that the two statements are not equivalent, noting that the left side is a number while the right side is a function, and clarifies that (lim x-->0) sin(x) = (lim x-->0) x results in 0 = 0.
  • A third participant points out that while the equality of limits is a necessary condition for the limit of the ratio to equal 1, it is not sufficient, providing an example involving sin(x) and x^2 to illustrate this point.

Areas of Agreement / Disagreement

Participants express disagreement regarding the equivalence of the limits, with some asserting that they are not the same and others questioning the implications of the limits in different contexts. The discussion remains unresolved regarding the sufficiency of the conditions for the limits.

Contextual Notes

Participants highlight the need for careful consideration of the definitions and conditions under which limits are evaluated, particularly in the context of functions approaching zero.

agapito
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From elementary calculus it is known that

(lim x-->0) ((sin x)/x) = 1.

Is this result equivalent to (lim x-->0) sin x = x ?

If so, how is it proved? Many thanks for all guidance.
 
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No, that makes no sense.
The left is a number, the limit of sin(x) as x goes to 0 (which happens to be 0) while the right is a function, x.

What is true is that $\lim_{x\to 0} sin(x)= \lim_{x\to 0} x$ which is simply 0= 0.
 
OK thanks for responding
 
I think I should point out that while $\lim_{x\to a} f(x)= \lim_{x\to a}g(x)$ is a necessary condition for $\lim_{x\to a}\frac{f(x)}{g(x)}= 1$ it is not sufficient.

For example $\lim_{x\to 0} sin(x)= \lim_{x\to 0} x^2= 0$ but $\lim_{x\to 0}\frac{sin(x)}{x^2}$ does not exist.
 

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