Sinx/x x tends to 0

  • Thread starter JG89
  • Start date
  • #1
726
1
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] ??
 

Answers and Replies

  • #2
256
0


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
726
1


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
lurflurf
Homework Helper
2,436
135


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 Threads on Sinx/x x tends to 0

Replies
6
Views
2K
Replies
14
Views
8K
Replies
8
Views
2K
Replies
2
Views
3K
Replies
13
Views
387
  • Last Post
Replies
2
Views
720
  • Last Post
Replies
7
Views
22K
  • Last Post
Replies
1
Views
1K
Replies
12
Views
2K
  • Last Post
Replies
2
Views
23K
Top