JEE Advanced 2022 · Paper 2 · Mathematics · Limits
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β=limx→0ex3−(1−x3)13+((1−x2)12−1)sinxx⋅sin2xx2⋅x2\beta = \lim_{x \to 0}\dfrac{{e}^{{x}^{3}} - \left(1 - {x}^{3}\right)\dfrac{1}{3} + \left(\left(1 - {x}^{2}\right)\dfrac{1}{2} - 1\right)\sin x}{x \cdot \dfrac{\sin^{2}x}{{x}^{2}} \cdot {x}^{2}}β=limx→0x⋅x2sin2x⋅x2ex3−(1−x3)31+((1−x2)21−1)sinxβ=limx→0((1+x3+x62!+⋯)−(1−13x3+(−19)x6+⋯)+(−12x2−18x4+⋯)(x−x33!+⋯)x3)\beta = \lim_{x \to 0}\left(\dfrac{\left(1 + {x}^{3} + \dfrac{{x}^{6}}{2 !} + ⋯\right) - \left(1 - \dfrac{1}{3}{x}^{3} + \left(- \dfrac{1}{9}\right){x}^{6} + ⋯\right) + \left(- \dfrac{1}{2}{x}^{2} - \dfrac{1}{8}{x}^{4} + ⋯\right)\left(x - \dfrac{{x}^{3}}{3 !} + ⋯\right)}{{x}^{3}}\right)β=limx→0x3(1+x3+2!x6+⋯)−(1−31x3+(−91)x6+⋯)+(−21x2−81x4+⋯)(x−3!x3+⋯)β=limx→0x3(1+13−12)x3\beta = \lim_{x \to 0}\dfrac{{x}^{3}\left(1 + \dfrac{1}{3} - \dfrac{1}{2}\right)}{{x}^{3}}β=limx→0x3x3(1+31−21) (Neglecting higher powers of xxx)β=56\beta = \dfrac{5}{6}β=65 i.e. 6β=56\beta = 56β=5
β=limx→0ex3−(1−x3)13+((1−x2)12−1)sinxx⋅sin2xx2⋅x2\beta = \lim_{x \to 0}\dfrac{{e}^{{x}^{3}} - \left(1 - {x}^{3}\right)\dfrac{1}{3} + \left(\left(1 - {x}^{2}\right)\dfrac{1}{2} - 1\right)\sin x}{x \cdot \dfrac{\sin^{2}x}{{x}^{2}} \cdot {x}^{2}}β=limx→0x⋅x2sin2x⋅x2ex3−(1−x3)31+((1−x2)21−1)sinx
β=limx→0((1+x3+x62!+⋯)−(1−13x3+(−19)x6+⋯)+(−12x2−18x4+⋯)(x−x33!+⋯)x3)\beta = \lim_{x \to 0}\left(\dfrac{\left(1 + {x}^{3} + \dfrac{{x}^{6}}{2 !} + ⋯\right) - \left(1 - \dfrac{1}{3}{x}^{3} + \left(- \dfrac{1}{9}\right){x}^{6} + ⋯\right) + \left(- \dfrac{1}{2}{x}^{2} - \dfrac{1}{8}{x}^{4} + ⋯\right)\left(x - \dfrac{{x}^{3}}{3 !} + ⋯\right)}{{x}^{3}}\right)β=limx→0x3(1+x3+2!x6+⋯)−(1−31x3+(−91)x6+⋯)+(−21x2−81x4+⋯)(x−3!x3+⋯)
β=limx→0x3(1+13−12)x3\beta = \lim_{x \to 0}\dfrac{{x}^{3}\left(1 + \dfrac{1}{3} - \dfrac{1}{2}\right)}{{x}^{3}}β=limx→0x3x3(1+31−21) (Neglecting higher powers of xxx)
β=56\beta = \dfrac{5}{6}β=65
i.e. 6β=56\beta = 56β=5
