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

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In summary, the conversation discusses different methods to prove that as x tends to 0, (sinx)/x tends to 1, with a focus on using L'Hospital's rule. However, it is pointed out that this rule is not needed and other methods, such as directly using the definition of the limit or geometric observations, can also be used. The conclusion is that the limit of (sinx)/x is 1 and different definitions and approaches can be used to prove this.
  • #1
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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  • #2


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!
 
  • #3


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 :)
 
  • #4


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
 

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

What is the definition of a limit?

The limit of a function at a particular point is the value that the function approaches as the independent variable approaches that point.

Why is the limit of sinx/x equal to 1?

This can be proven using the squeeze theorem, which states that if two functions have the same limit at a point and a third function is between them, then the third function also has the same limit at that point.

What is the significance of the limit of sinx/x equal to 1?

This limit is significant because it is a fundamental result in calculus and is used in various applications, such as finding derivatives and integrals.

Can the limit of sinx/x be proven using other methods?

Yes, it can also be proven using L'Hopital's rule, which states that the limit of the quotient of two differentiable functions is equal to the limit of the quotient of their derivatives.

Are there any exceptions to the limit of sinx/x being equal to 1?

Yes, there are some cases where the limit may not be equal to 1, such as when x approaches infinity or when x is a multiple of π.

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