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  • Confused! kindly explain, - - Limit , continuity and differentiability - JEE Main

\lim_{y\rightarrow 0} \frac{\sqrt{1+\sqrt{1+y^4}}- \sqrt2}{y^4}

  • Option 1)

    exists and equals \frac{1}{4\sqrt2}

  • Option 2)

    exists and equals \frac{1}{2\sqrt2 (\sqrt2 + 1)}

  • Option 3)

    exists and equals \frac{1}{2\sqrt2}

  • Option 4)

    does not exist

Answers (1)

best_answer

 

Angle of intersection of two curves -

The angle of intersection of two curves is the angle subtended between the tangents at their point of intersection.Let  m1  &  m2 are two slope of tangents at intersection point of two curves then

tan\theta=\frac{[m_{1}-m_{2}]}{1+m_{1}m_{2}}

- wherein

where \theta is angle between two curves tangents.

\lim_{y\rightarrow 0}\frac{\sqrt{1+\sqrt{1+r^{4}}}-\sqrt{2}}{y^{4}}

factorize this 

\Rightarrow \lim_{y\rightarrow 0}\frac{\sqrt{1+\sqrt{1+y^{4}}}-\sqrt{2}}{y^{4}}\times \frac{\sqrt{1+\sqrt{1+y^{4}}}+\sqrt{2}}{\sqrt{1+\sqrt{1+y^{4}}}+\sqrt{2}}

\Rightarrow \lim_{y\rightarrow 0}\frac{1+\sqrt{1+y^{4}}-2}{y^{4}\left ( \sqrt{1+\sqrt{1+y^{4}}} \right )+\sqrt{2}}

again factorize

\Rightarrow \lim_{y\rightarrow 0}\frac{\left ( \sqrt{1+y^{4}} \right )-1}{y^{4}\left ( \sqrt{1+\sqrt{1+y^{4}}} \right )+\sqrt{2}}\times \frac{\sqrt{1+y^{4}}+1}{\sqrt{1+y^{4}}+1}

\Rightarrow \lim_{y\rightarrow 0}\frac{1+y^{4}-1}{y^{4}\left ( \sqrt{1+\sqrt{1+y^{4}}+\sqrt{2}} \right )\left ( \sqrt{1+y^{4}}+1 \right )}

\Rightarrow \lim_{y\rightarrow 0}\frac{1}{\left ( \sqrt{1+\sqrt{1+y^{4}}+\sqrt{2}} \right )\left ( \sqrt{1+y^{4}}+1 \right )}=\frac{1}{4\sqrt{2}}

 


Option 1)

exists and equals \frac{1}{4\sqrt2}

Option 2)

exists and equals \frac{1}{2\sqrt2 (\sqrt2 + 1)}

Option 3)

exists and equals \frac{1}{2\sqrt2}

Option 4)

does not exist

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