Question

1-cos^3x
lim
x->0 x sin 2x​

Answer 1

Answer:

$\dfrac{3}{4} \: is \: the \: answer$

Step-by-step explanation:

We Have :-

$lim_{x - > 0} \: \dfrac{1 - cos^{3}x}{xsin2x}$

Solution :-

$lim_{x - > 0} \: \dfrac{1 - cos^{3}x}{x \: sin2x} \\ \\ \\ applying \: l \: hopital \: rule \\ \\ \\ lim_{x - > 0} \: \dfrac{3 \: cos^{2}x \: sinx }{2x \: cos2x + sin2x } \\ \\ \\ lim_{x - > 0} \: \dfrac{3 \: cos^{2}x \: sinx }{2x \: (cos^{2}x \: - sin^{2} x) + 2 \: sinx \: cosx } \\ \\ \\ \dfrac{3}{2} lim_{x - > 0} \: \dfrac{\: cos^{2}x \: sinx }{x \: (cos^{2}x \: - sin^{2} x) + \: sinx \: cosx } \\ \\ \\ dividing \: by \: cos^{2} x \\ \\ \\ \dfrac{3}{2} lim_{x - > 0} \: \dfrac{ sinx }{x \: (1 - tan^{2} x) + tanx } \\ \\ \\ again \: applying \: l \: hopital \\ \\ \\ \dfrac{3}{2} lim_{x - > 0} \: \dfrac{ cosx }{x \: ( - 2tanx \: sec^{2} x) + (1 - tan^{2}x) \times 1 + sec^{2} x} \\ \\ \\ putting \: the \: limit \: \\ \\ \\ \dfrac{3}{2} \times \dfrac{1}{1 + 1} \\ \\ \\ \dfrac{3}{2} \times \dfrac{1}{2} \\ \\ \\ \dfrac{3}{4}$

Answer 2

$\huge{\text{\underline{Solution:-}}}$

$\implies$Lim( x→0) ( 1- cos³x) / xsin2x

★ Now,

$\implies$(1 - cos³x) = (1 - cosx) (1 + cos²x + cosx)

$\implies$(2sin²x / 2) (1 + cos²x + cosx)

★ Using concept:-

$\implies$Lim( x → 0) sinx / x = 1

★ Now,

$\implies$Lim ( x →0) (2sin²x / 2 / x²/ 4 × x²/ 4 ) ( 1 + cos²x + cosx) / x (sin2x / 2x) × 2x

$\implies$ ( 2 × 1 / 4) ( 1 + 1 + 1) / ( 2)

$\implies\tt{\boxed{\frac{3}{4}}}$

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