Question

how to prove that lim x->0 (sin x)/x =1

Answer 1

i remember asking this problem once...you can look at the explanation they gave me!

http://openstudy.com/users/lgbasallote#/updates/4e8144da0b8bc11dd53dd8c1

Answer 2

\[\lim_{x \rightarrow 0}(\sin x) \div x\]

Answer 3

We know from the range of sin(1/x)
-1 <= sin(1/x) <= 1

multiplying through by x, you obtain [note see edit below about this]
-x <= x*sin(1/x) <= x

taking the limit as x->0 we get
0 <= lim x->0 x*sin(1/x) <= 0

therefore by squeeze/sandwich theorem the
lim x->0 x*sin(1/x) =0

edit: starwhitedwarf: excellent point.... I should have used |x| like sahsjing and yourself, or did cases.. With cases:

-x <= x*sin(1/x) <= x for x>=0
-x > x*sin(1/x) > x for x< 0

Then right and left handed limits
0 <= lim x->0+ x*sin(1/x) <= 0
0 > lim x->0- x*sin(1/x) > 0

Answer 4

do you know the L'Hopital's rule? well using the L'Hoptial's rule, we know that the limit of this is the same as...
\[\lim_{x \rightarrow 0}cosx\]
=1

Answer 5

Easy proof:
sinx<x<tgx for 0<x<Pi/2
no devide sinx by every member of previous enequality (remmeber 0<x<Pi/2)
we get:
1>sinx/x>cosx
multiply by -1 and rearange a bit:
0<1-sinx/x<1-cosx
but \[1-cosx =2\sin ^{2}x/2<2sinx/2<x\]
so
0<1-sinx/x<x
as x goes to 0 you get your proof