Limits

Calculus Level pending

lim x 0 sin x x \lim _{ x\rightarrow 0 }{ \frac { \sin { x } }{ x } }

1 2 \frac { 1 }{ \sqrt { 2 } } π \pi 0 1

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1 solution

Rajdeep Dhingra
Nov 4, 2014

U s i n g E u l e r s F o r m u l a w e g e t sin x = x x 3 3 ! + x 5 5 ! + H . O . T H . O . T = h i g h e r o r d e r t e r m s P u t t h i s i n e q u sin x x = 1 x 2 3 ! + x 4 5 ! + H . O . T p u t t i n g x = 0 w e g e t lim x 0 sin x x = 1 Using\quad Euler's\quad Formula\quad we\quad get\\ \sin { x } =\quad x\quad -\quad \frac { { x }^{ 3 } }{ 3! } \quad +\quad \frac { { x }^{ 5 } }{ 5! } \quad +\quad H.O.T\\ H.O.T\quad =\quad higher\quad order\quad terms\\ Put\quad this\quad in\quad equ\\ \frac { \sin { x } }{ x } =\quad 1\quad -\quad \frac { { x }^{ 2 } }{ 3! } \quad +\quad \frac { { x }^{ 4 } }{ 5! } \quad +\quad H.O.T\\ putting\quad x\quad =\quad 0\quad we\quad get\\ \lim _{ x\rightarrow 0 }{ \frac { \sin { x } }{ x } } \quad =\quad 1

Proof of sinx Proof of sinx Any doubts please Reply

Rajdeep Dhingra - 6 years, 7 months ago

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