anonymous
  • anonymous
how to prove that lim x->0 (sin x)/x =1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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lgbasallote
  • lgbasallote
i remember asking this problem once...you can look at the explanation they gave me! http://openstudy.com/users/lgbasallote#/updates/4e8144da0b8bc11dd53dd8c1
anonymous
  • anonymous
\[\lim_{x \rightarrow 0}(\sin x) \div x\]
anonymous
  • anonymous
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

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anonymous
  • anonymous
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
anonymous
  • anonymous
Easy proof: sinxsinx/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

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