How do you find $lim_(theta->0)$ $((1-cos^2 theta)/theta ^2)$ ?

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Answer 1 · 2018-05-13T09:33:34

$= 1$

We know $1 - cos^2 theta = sin^2 theta$

$=> lim_( theta to 0 ) sin^2 theta / theta^2$

$=> lim_(theta to 0 ) (sin theta / theta )^2$

Use an approximation for $sin theta$

$=> sin theta = theta - theta^3 / (3! )$

$=> lim_( theta to 0 )( ( theta - theta^3/(3! ) ) / theta ) ^2$

$=> lim_ ( theta to 0 ) (1 - theta^2 / (3! ))^2$

$= 1$

Answer 2 · 2018-05-16T07:42:26

The limit is $=1$

We need

$lim_(theta->0)(1-cos^2theta)/theta^2=(1-1)/0=0/0$

which is indeterminate, so apply L'Hopital's rule

$lim_(theta->0)(1-cos^2theta)/theta^2=lim_(theta->0)((1-cos^2theta)')/((theta^2)')$

$=lim_(theta->0)(2costhetasintheta)/(2theta)$

$=lim_(theta->0)(sin2theta)/(2theta)$

$=lim_(theta->0)((sin2theta)')/((2theta)')$

$=lim_(theta->0)(2cos2theta)/(2)$

$=1$

How do you find lim_(theta->0)lim_(theta->0) ((1-cos^2 theta)/theta ^2)((1-cos^2 theta)/theta ^2)?