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  • How to solve this problem- - Limit , continuity and differentiability - JEE Main

\lim_{x\rightarrow 0}\frac{e^{x^{2}}-\cos \! x}{\sin ^{2}x}   is equal to :

 

  • Option 1)

    3

  • Option 2)

    \frac{3}{2}

  • Option 3)

    \frac{5}{4}

  • Option 4)

    2

 

Answers (1)

best_answer

As we learnt in 

Condition for Trigonometric limit -

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

It\:must\:be\:\frac{sin(angle)}{(angle)_{(in\:radian)}}

- wherein

ex:-

\lim_{x\rightarrow 0}\:\:\frac{sinx}{x}=\frac{\pi}{180}

 

 =>\lim_{x\to 0} \frac{e^{x^{2}}-\cos x}{sin^{2}x}

=>\lim_{x\to 0}\frac{e^{x^{2}}\times 2x+\sin x}{2\sin x\cos x}

=>\lim_{x\to 0}\frac{e^{x^{2}}\times 2+\frac{\sin x}{x}}{2\left ( \frac{\sin x}{x} \right )\cos x}

=>\frac{2\times 1+1}{2\times 1}

=\frac{3}{2}


Option 1)

3

Incorrect option    

Option 2)

\frac{3}{2}

Correct option

Option 3)

\frac{5}{4}

Incorrect option    

Option 4)

2

Incorrect option    

Posted by

divya.saini

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