# Sinx/x x tends to 0

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

$$\lim_{x \to 0} x$$ 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!!!

$$\lim_{x \to 0} x$$ 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!!!
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).

lurflurf
Homework Helper

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