\lim_{x\rightarrow 0}\frac{\sin^{2}x}{\sqrt{2}-\sqrt{1+\cos\, x}} equals :
 

  • Option 1)

    4\sqrt{2}

  • Option 2)

    \sqrt{2}

  • Option 3)

    2\sqrt{2}

     

  • Option 4)

    4

 

Answers (1)

\lim_{x\rightarrow 0}\frac{\sin^{2}x}{\sqrt{2}-\sqrt{1+\cos\: x}}

normalizing :

: \frac{(\sin^{2}x(\sqrt{2}+\sqrt{1+\cos x})}{(2-1-\cos x)}

=(\sin^{2}x)=1-\cos^{2}x=(1-\cos x)(1+\cos\; 7)

\lim_{x\rightarrow 0}\frac{(1-cos\, x)(1+\cos x)(\sqrt{2}+\sqrt{1+cos x})}{(1-cos\, x)}

=\lim_{x\rightarrow 0}\left ( 1+cos\, x \right )(\sqrt{2}+\sqrt{1+cos\, x})

=(1+1)(\sqrt{2}+\sqrt{1+1})

=2(2\sqrt{2})

=4\sqrt{2}


Option 1)

4\sqrt{2}

Option 2)

\sqrt{2}

Option 3)

2\sqrt{2}

 

Option 4)

4

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