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$\lim_{x\rightarrow 2}\left ( \frac{\sqrt{1-\cos \left \{ 2(x-2) \right \}}}{x-2} \right )$

Option 1)
$equals-\sqrt{2}\;$

Option 2)
$\; \; equals\frac{1}{\sqrt{2}}\;$

Option 3)
does not exist

Option 4)
$\; equals\sqrt{2}\;$

Answers (1)

best_answer

As we learnt in

Evalution of Trigonometric limit -

$\lim_{x\rightarrow a}\:\frac{sin(x-a)}{x-a}=1$

$\lim_{x\rightarrow a}\:\frac{tan(x-a)}{x-a}=1$

$put\:\:\:\:\:x=a+h\:\:\:where\:\:h\rightarrow 0$

$Then\:it\:comes$

$\lim_{h\rightarrow 0}\:\:\frac{sinh}{h}=\lim_{h\rightarrow 0}\:\:\frac{tanh}{h}=1$

$\therefore\:\:\:\lim_{x\rightarrow 0}\:\:\frac{sinx}{x}=1\:\;\;and$

$\therefore\:\:\:\lim_{x\rightarrow 0}\:\:\frac{tanx}{x}=1$

-

$\lim_{x\rightarrow 2}\:\frac{\sqrt{1-cos(2(x-2))}}{(x-2)}$

$\lim_{x\rightarrow 2}\:\frac{\sqrt{2sin^{2}(x-2)}}{(x-2)}$

$\lim_{x\rightarrow 2}\:\sqrt{2}\:\frac{[sin(x-2)]}{x-2}$

For x-2 $> 0$ it is $\sqrt{2}$

For x-2 $< 0$ it is $-\sqrt{2}$

So LHL $\neq$ PHL

So limit does not exist.

Option 1)

$equals-\sqrt{2}\;$

This option is incorrect.

Option 2)

$\; \; equals\frac{1}{\sqrt{2}}\;$

This option is incorrect.

Option 3)

does not exist

This option is correct.

Option 4)

$\; equals\sqrt{2}\;$

This option is incorrect.

Posted by

divya.saini

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