Sinx/x x tends to 0

1. Jan 28, 2009

JG89

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:
$$\lim_{x \rightarrow 0} x = \lim_{x \rightarrow 0} sinx$$ and so $$\lim_{x \rightarrow 0} \frac{x}{x} = 1 = \lim_{x \rightarrow 0} \frac{sinx}{x}$$ ??

2. Jan 28, 2009

rochfor1

Re: Sinx/x

$$\lim_{x \to 0} x$$ by writing $$x=\frac{ x^2 }{ x }$$. Also the limit of a quotient is only the quotient of the limits when the bottom limit is nonzero!

Use L'Hospital!!!

3. Jan 28, 2009

JG89

Re: Sinx/x

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.

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

4. Jan 29, 2009

lurflurf

Re: Sinx/x

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