Limx➡️0,x^2/sinx^2don't spam – InterviewSolution

1.

Limx➡️0,x^2/sinx^2don't spam

Answer»

GIVEN :–

• Limit –

$\\ \implies \bf\lim_{x \to0} \bigg[\dfrac{ {x}^{2} }{ \sin( {x}^{2} ) } \bigg] \\$

TO FIND :–

• Value of limit = ?

SOLUTION :–

• Let –

$\\ \implies \bf P = \lim_{x \to0} \bigg[\dfrac{ {x}^{2} }{ \sin( {x}^{2} ) } \bigg] \\$

• PUT limit –

$\\ \implies \bf P = \bigg[\dfrac{ {(0)}^{2} }{ \sin( {0}) } \bigg] \\$

$\\ \implies \bf P =\dfrac{0}{0 } \\$

• It's undefined form. So let's use L'HOSPITAL RULE –

$\\ \implies \bf P = \lim_{x \to0} \bigg[\dfrac{2x}{2x \cos( {x}^{2} ) } \bigg] \\$

$\\ \implies \bf P = \lim_{x \to0} \bigg[\dfrac{ \cancel{(2x)}}{ \cancel{(2x)}\cos( {x}^{2} ) } \bigg] \\$

$\\ \implies \bf P = \lim_{x \to0} \bigg[\dfrac{1}{\cos( {x}^{2} ) } \bigg] \\$

• Again put limit –

$\\ \implies \bf P = \bigg[\dfrac{1}{\cos(0) } \bigg] \\$

$\\ \implies \bf P = \bigg[\dfrac{1}{1} \bigg] \\$

$\\ \implies \large \red { \boxed{ \bf P =1}}\\$

• HENCE –

$\\ \implies\bf \lim_{x \to0} \bigg[\dfrac{ {x}^{2} }{ \sin( {x}^{2} ) } \bigg]=1 \\$

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