Verified Solution • 2025–26 Session moderate QJEEM02Maf941: JEE Main 2002 Mathematics - Class 12 JEEM02Maf941 Solution limx→∞(x2+5x+3x2+x+2)x\mathop {\lim }\limits_{x \to \infty } {\left( {{{{x^2} + 5x + 3} \over {{x^2} + x + 2}}} \right)^x}x→∞lim(x2+x+2x2+5x+3)x Multiple Choice Options A e4{e^4}e4 B e2{e^2}e2 C e3{e^3}e3 D 111 Step-by-Step Verified Solution limx→∞(x2+5x+3x2+x+2)x\mathop {\lim }\limits_{x \to \infty } {\left( {{{{x^2} + 5x + 3} \over {{x^2} + x + 2}}} \right)^x}x→∞lim(x2+x+2x2+5x+3)x =limx→∞(1+4x+1x2+x+2)x= \mathop {\lim }\limits_{x \to \infty } {\left( {1 + {{4x + 1} \over {{x^2} + x + 2 }}} \right)^x}=x→∞lim(1+x2+x+24x+1)x =limx→∞[(1+4x+1x2+x+2)x2+x+24x+1](4x+1)xx2+x+2= \mathop {\lim }\limits_{x \to \infty } {\left[ {{{\left( {1 + {{4x + 1} \over {{x^2} + x + 2}}} \right)}^{{{{x^2} + x + 2} \over {4x + 1}}}}} \right]^{{{\left( {4x + 1} \right)x} \over {{x^2} + x + 2}}}}=x→∞lim[(1+x2+x+24x+1)4x+1x2+x+2]x2+x+2(4x+1)x = limx→∞[(1+4x+1x2+x+2)14x+1x2+x+2](4x+1)xx2+x+2\mathop {\lim }\limits_{x \to \infty } {\left[ {{{\left( {1 + {{4x + 1} \over {{x^2} + x + 2}}} \right)}^{{1 \over {{{4x + 1} \over {{x^2} + x + 2}}}}}}} \right]^{{{\left( {4x + 1} \right)x} \over {{x^2} + x + 2}}}}x→∞lim[(1+x2+x+24x+1)x2+x+24x+11]x2+x+2(4x+1)x =elimx→∞4x2+xx2+x+2= {e^{\mathop {\lim }\limits_{x \to \infty } {{4{x^2} + x} \over {{x^2} + x + 2}}}}=ex→∞limx2+x+24x2+x [ As limx→∞(1+λx)1x=eλ\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \lambda x} \right)^{{1 \over x}}} = {e^\lambda }x→∞lim(1+λx)x1=eλ ] =elimx→∞4+1x1+1x+2x2=e4= {e^{\mathop {\lim }\limits_{x \to \infty } {{4 + {1 \over x}} \over {1 + {1 \over x} + {2 \over {{x^2}}}}}}} = {e^4}=ex→∞lim1+x1+x224+x1=e4 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring