Question

evaluate lim x→0 x^2+2cosx-2/xsin^3x​

Answer 1

$\LARGE\bf{\underline{\underline{\pink{\bigstar Answer}}}}$

$\large\rm{ \lim\limits_{x \to 0} \Bigg ( \dfrac{ x^2 + 2 \ \cos \ x -2}{ x \ \sin^{3} x} \Bigg )}$

$\large\rm$

$\large\rm{ = \lim\limits_{x \to 0} \Bigg ( \dfrac{ x^2 + 2 \ \cos \ x -2}{ x^{4}} \times \dfrac{x^{3} }{ \sin^{3} \ x} \Bigg )}$

$\large\rm$

$\large\rm{ = \lim\limits_{x \to 0} \Bigg ( \dfrac{ x^2 + 2 \ \cos \ x -2}{ x^{4}} \Bigg )}$

$\large\rm$

$\large\rm{ = \lim\limits_{ x \to 0} \Bigg ( \dfrac{2x - 2 \sin \ x}{4x^{3}} \Bigg ) }$

$\large\rm$

$\large\rm{ = \lim\limits_{ x \to 0} \Bigg ( \dfrac{2x - 2 \cos \ x}{12x^{3}} \Bigg ) }$

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$\large\rm{ = \lim\limits_{x \to 0} \Bigg ( \dfrac{2 \sin \ x}{24 x^3} \Bigg ) }$

$\large\rm$

$\large\rm{ = \dfrac{1}{2}}$

Answer 2

Answer:

lim(x sin3xx2+2 cos x−2)

\large\rm

\large\rm{ = \lim\limits_{x \to 0} \Bigg ( \dfrac{ x^2 + 2 \ \cos \ x -2}{ x^{4}} \times \dfrac{x^{3} }{ \sin^{3} \ x} \Bigg )}=x→0lim(x4x2+2 cos x−2×sin3 xx3)

\large\rm

\large\rm{ = \lim\limits_{x \to 0} \Bigg ( \dfrac{ x^2 + 2 \ \cos \ x -2}{ x^{4}} \Bigg )}=x→0lim(x4x2+2 cos x−2)

\large\rm

\large\rm{ = \lim\limits_{ x \to 0} \Bigg ( \dfrac{2x - 2 \sin \ x}{4x^{3}} \Bigg ) }=x→0lim(4x32x−2sin x)

\large\rm

\large\rm{ = \lim\limits_{ x \to 0} \Bigg ( \dfrac{2x - 2 \cos \ x}{12x^{3}} \Bigg ) }=x→0lim(12x32x−2cos x)

\large\rm

\large\rm{ = \lim\limits_{x \to 0} \Bigg ( \dfrac{2 \sin \ x}{24 x^3} \Bigg ) }=x→0lim(24x32sin x)

\large\rm

\large\rm{ = \dfrac{1}{2}}=21