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\( \lim _{x \rightarrow 0} \frac{\tan (\sin x)-x}{\tan x^{3}} \) is equal to (1) \( \frac{1}{6} \) (2) \( \frac{1}{3} \)(3) \( \frac{1}{2} \) (4) 1 |
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Answer» \(\lim\limits_{x \to 0} \) \(\cfrac{tan(sinx)-x}{tanx^3}\) \(\)(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) = \(\cfrac13\) \(\therefore\) \(\lim\limits_{x \to 0} \) \(\cfrac{tan(sinx)-x}{tanx^3}\) = \(\cfrac13\) |
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