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  • Solve this problem - Limit , continuity and differentiability - JEE Main-3

\lim_{x\rightarrow 0}\frac{\sqrt{1-\cos 2x}}{\sqrt{2}x} is

  • Option 1)

    1

  • Option 2)

    -1

  • Option 3)

    0

  • Option 4)

    does not exist.

 

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 0} \frac{\sqrt{1-cos2x}}{x\sqrt{2}}

\Rightarrow \lim_{x\rightarrow 0} \frac{\sqrt{2sin^2x}}{x\sqrt{2}}

\Rightarrow \lim_{n\rightarrow 0} \frac{\sqrt{2}}{\sqrt{2}}.\frac{ |sinx|}{x}

for x\rightarrow 0^{+} it is 1

for x\rightarrow 0^{-} it is -1

So limit does not exist.


Option 1)

1

Correct

Option 2)

-1

Incorrect

Option 3)

0

Incorrect

Option 4)

does not exist.

Incorrect

Posted by

Vakul

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