lim x-->0 {[tan(sinx)-x]/tanx^(3)} is equal to?

Question

\( \lim _{x \rightarrow 0} \frac{\tan (\sin x)-x}{\tan x^{3}} \) is equal to 

(1) \( \frac{1}{6} \) 

(2) \( \frac{1}{3} \)

(3) \( \frac{1}{2} \) 

(4) 1

Answer 1

\(\lim\limits_{x \to 0} \) \(\cfrac{tan(sinx)-x}{tanx^3}\) \(\)(0/0 - type)

= \(\lim\limits_{x \to 0} \) \(\cfrac{sec^2(sinx).cosx-1}{sec^2(x^3).3x^2}\) (BY D.L.H. Rule)

= \(\cfrac{\lim\limits_{x\to 0}sec^2sinx\lim\limits_{x\to 0}cosx-1}{\lim\limits_{x\to 0}sec^2x^3.\lim\limits_{x\to 0}3x^2}\)

= \(\lim\limits_{x \to 0} \) = \(\cfrac{sec^2sinx-1}{3x^2}\) (\(\because\) \(\lim\limits_{x \to 0} \) cosx = 1 and \(\lim\limits_{x \to 0} \) sec²x³ = 1)

(0/0. case)=

= \(\lim\limits_{x \to 0} \)\(\cfrac{2sec^2(sinx)tan(sinx).cosx}{6x}\)

= \(\cfrac26\) \(\lim\limits_{x \to 0} \) sec² (sinx) . cosx \(\lim\limits_{x \to 0} \) \(\cfrac{tan(sinx)}x\) (0/0 type)

=  \(\cfrac13\) sec² 0. cos 0 \(\lim\limits_{x \to 0} \) sec² (sinx) cos x

= \(\cfrac13\) x 1 x 1 sec² 0 . cos 0

= \(\cfrac13\) x 1 x 1 = \(\cfrac13\) 

Alternative = \(\lim\limits_{x \to 0} \) \(\cfrac{tan(sinx)-x}{tanx^3}\) 

= \(\lim\limits_{x \to 0} \) \(\cfrac{tanx-x}{x^3}\) (0/0case)(\(\because\) \(\theta\) is very small then sin \(\theta\) ≈ \(\theta\) and tan   \(\theta\) ≈ \(\theta\) )

= \(\lim\limits_{x \to 0} \) \(\cfrac{sec^2x-1}{3x^2}\) (0/0case)(By  D.L.H Rule)

= \(\lim\limits_{x \to 0} \) \(\cfrac{2sec^2xtanx}{6x}\) 

= \(\cfrac26\) \(\lim\limits_{x \to 0} \) sec² x \(\lim\limits_{x \to 0} \) \(\cfrac{tanx}x\) 

= \(\cfrac13\) x sec² 0 x 1 (\(\because\) \(\lim\limits_{x \to 0} \) \(\cfrac{tanx}x\) = 1)

= \(\cfrac13\) 

\(\therefore\) \(\lim\limits_{x \to 0} \) \(\cfrac{tan(sinx)-x}{tanx^3}\) = \(\cfrac13\)