First slide

Introduction to limits

Question

$lim_{x \to 3} \frac{\sqrt{1 - \cos 2 (x - 3)}}{x - 3}$

Easy

A

$= \sqrt{2}$
B

does not exist
C

=1
D

$= - \sqrt{2}$

Solution

$1 - \cos 2 (x - 3) = 2 \sin^{2} (x - 3)$
$\sqrt{1 - \cos 2 (x - 3)} = \sqrt{2} | \sin (x - 3) |$
so, $lim_{x \to 3 +} \frac{\sqrt{1 - \cos 2 (x - 3)}}{x - 3} = \sqrt{2} lim_{x \to 3 +} \frac{\sin (x - 3)}{x - 3} = \sqrt{2}$
where as
$lim_{x \to 3 -} \frac{\sqrt{1 - \cos 2 (x - 3)}}{x - 3} = \sqrt{2} lim_{x \to 3 -} \frac{- \sin (x - 3)}{x - 3} = - \sqrt{2}$

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