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QJEEM03Ma2005: JEE Main 2003 Mathematics - Class 12 JEEM03Ma2005 Solution

The value of lim⁡x→0∫0x2sec⁡2tdtxsinx\mathop {\lim }\limits_{x \to 0} {{\int\limits_0^{{x^2}} {{{\sec }^2}tdt} } \over xsinx} is

Multiple Choice Options

  • A 0
  • B 3
  • C 2
  • D 1

Step-by-Step Verified Solution

lim⁡x→0ddx∫0x2sec⁡2tdtdx(xsin⁡x)\mathop {\lim }\limits_{x \to 0} {{{d \over {dx}}\int\limits_0^{{x^2}} {{{\sec }^2}tdt} } \over {{d \over x}\left( {x\sin x} \right)}}

=lim⁡x→0sec⁡2x2.2xsin⁡ x+x cos⁡ x= \mathop {\lim }\limits_{x \to 0} {{{{\sec }^2}{x^2}.2x} \over {\sin \,x + x\,\cos \,x}} (by L′L' Hospital rule)

lim⁡x→02sec⁡2x2(sin⁡xx+cos⁡ x)=2×11+1=1\mathop {\lim }\limits_{x \to 0} {{2{{\sec }^2}{x^2}} \over {\left( {{{\sin x} \over x} + \cos \,x} \right)}} = {{2 \times 1} \over {1 + 1}} = 1

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