The value of mathop {lim }limits_{a to 0} fra | vedclass

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Question

(A) We are given the limit: $\mathop {\lim }\limits_{a 	o 0} \frac{{\sin a - 	an a}}{{{{\sin }^3}a}}$Rewrite $	an a$ as $\frac{\sin a}{\cos a}$:$\m

Vedclass · Accepted Answer

$ - \frac{1}{2}$

Vedclass · Answer

(A) We are given the limit: $\mathop {\lim }\limits_{a 	o 0} \frac{{\sin a - 	an a}}{{{{\sin }^3}a}}$Rewrite $	an a$ as $\frac{\sin a}{\cos a}$:$\mathop {\lim }\limits_{a 	o 0} \frac{{\sin a - \frac{\sin a}{\cos a}}}{{{{\sin }^3}a}} = \mathop {\lim }\limits_{a 	o 0} \frac{{\sin a(\cos a - 1)}}{{\cos a \cdot \sin^3 a}}$Simplify by canceling $\sin a$:$= \mathop {\lim }\limits_{a 	o 0} \frac{{\cos a - 1}}{{\cos a \cdot \sin^2 a}}$Use the identity $\sin^2 a = 1 - \cos^2 a = (1 - \cos a)(1 + \cos a)$:$= \mathop {\lim }\limits_{a 	o 0} \frac{{-(\cos a - 1)}}{{\cos a \cdot (1 - \cos a)(1 + \cos a)}}$Cancel $(1 - \cos a)$:$= \mathop {\lim }\limits_{a 	o 0} \frac{{-1}}{{\cos a(1 + \cos a)}}$Substitute $a = 0$:$= \frac{{-1}}{{1(1 + 1)}} = -\frac{1}{2}$

The value of $\mathop {\lim }\limits_{a \to 0} \frac{{\sin a - \tan a}}{{{{\sin }^3}a}}$ will be

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