Proof that $\lim_{x \rightarrow 0} \frac{sinx}{x} = 1$

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Discussion Overview

The discussion revolves around the proof of the limit $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$. Participants explore various methods and approaches to establish this limit, including direct evaluation, L'Hospital's rule, and definitions involving Taylor series and geometric interpretations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that since $\lim_{x \rightarrow 0} x = \lim_{x \rightarrow 0} \sin x$, it follows that $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$, but this is challenged.
  • Another participant argues against the initial claim, emphasizing that the limit of a quotient is valid only when the denominator is non-zero and suggests using L'Hospital's rule instead.
  • A later reply acknowledges the mistake in the initial reasoning and provides a counter-example to illustrate the point about limits of quotients.
  • Several definitions are presented to compute the limit directly, including trigonometric identities, Taylor series expansions, and geometric interpretations, each concluding with $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$.
  • One participant questions the frequent use of L'Hospital's rule, suggesting that it may not be necessary for this limit and proposes alternative definitions to derive the limit directly.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method to prove the limit. There are competing views on the necessity of L'Hospital's rule and the validity of different approaches to the proof.

Contextual Notes

Some definitions and methods presented rely on specific assumptions about the behavior of functions near zero, and the discussion highlights the importance of conditions under which certain limit properties hold.

JG89
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I know that as x tends to 0, (sinx)/x tends to 1. A post from GibZ got me thinking, would this be a proper proof of that:
[tex]\lim_{x \rightarrow 0} x = \lim_{x \rightarrow 0} sinx[/tex] and so [tex]\lim_{x \rightarrow 0} \frac{x}{x} = 1 = \lim_{x \rightarrow 0} \frac{sinx}{x}[/tex] ??
 
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No! Think about using your method to find
[tex]\lim_{x \to 0} x[/tex] by writing [tex]x=\frac{ x^2 }{ x }[/tex]. Also the limit of a quotient is only the quotient of the limits when the bottom limit is nonzero!

Use L'Hospital!
 


rochfor1 said:
No! Think about using your method to find
[tex]\lim_{x \to 0} x[/tex] by writing [tex]x=\frac{ x^2 }{ x }[/tex]. Also the limit of a quotient is only the quotient of the limits when the bottom limit is nonzero!

Use L'Hospital!

Ah! That's the part I forgot! That the limit of a quotient is a quotient of limits only when the denominator isn't 0. I constructed a counter-example too to prove myself wrong in the previous post.

[tex]\lim_{x \rightarrow 0} x = 0[/tex] and so [tex]\lim_{x \rightarrow 0} \frac{x}{x} = 1 = \lim_{x \rightarrow 0} \frac{0}{x} = 0[/tex] which obviously isn't true (because the denominator is tending to 0).

Thanks for the reply :)
 


rochfor1 said:
Use L'Hospital!


Why do people want to use L'Hospital on every problem? Is it because it is fun to say? L'Hospital is not needed.
lim sin(x)/x=lim [sin(0+x)-sin(x)]/[(0+x)-0]=sin'(0)=1

We might want to compute the limit directly from the definition. For that we need to choose a definition.

def1:
sin(x+y)=sin(x)cos(y)+cos(x)sin(y)
cos(x+y)=cos(x)cos(y)-sin(x)sin(y)
cos(x)^2+sin(x)^2=1
lim sin(x)/x=0

here the limit is included in the definition
QED

def2:
sin(x)=x-x^3/6+...

we can write
sin(x)=sin(x)+sin'(0) x+sin''(0) x^2/2+sin'''(0) x^3/6+...
equating like terms yields
sin'(0)=1
QED

def3:
sin''(x)+sin(x)=0
sin(0)=0
sin'(0)=1

sin'(0)=1 is in the definition
QED

def4:
sin(x)/x=(1-[x/(1pi)]^2)(1-[x/(2pi)]^2)(1-[x/(3pi)]^2)(1-[x/(4pi)]^2)(1-[x/(5pi)]^2)...

lim sin(x)/x=(1-[0/(1pi)]^2)(1-[0/(2pi)]^2)(1-[0/(3pi)]^2)(1-[0/(4pi)]^2)(1-[0/(5pi)]^2)...
=1
QED

def5:
various geometric junk from which it is noticed

for x small
sin(2x)/(2x)<=cos(x)<=sin(x)/x<=1
from which we conclude
cos(0)<=lim sin(x)/x<=1
lim sin(x)/x=1
QED
 

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