\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)

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.

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