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$\lim\limits_{x\to 0}\large\frac{\tan x-\sin x}{x^3}$=

$(a)\;1\qquad(b)\;2\qquad(c)\;\large\frac{1}{2}$$\qquad(d)\;-1$

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Toolbox:
  • L. Hospital,s Rule: $\lim\limits_{x\to 0}f(x)$=$\lim\limits_{x\to 0} f' (x)$
  • $\lim\limits_{x\to 0}\large\frac{f(x)}{g(x)}$=$\lim\limits_{x\to 0}\large\frac{f'(x)}{g'(x)}$ where $g(x)\neq0$
$\lim\limits_{x\to 0}\large\frac{\tan x-\sin x}{x^3}$
Let $f(x)=\tan x-\sin x$ and $g(x)=x^3$
$f'(x)=\sec^2 x-\cos x$ and $g'(x)=3x^3$
$\large \frac{f'(x)}{g'(x)}$=$\large\frac{\sec^2 x-\cos x}{3x^2}$
$\:\:\:=\large\frac{1-\cos^3 x}{3x^2.\;\cos^2 x}$ $[since \:\sec^2 x=\large\frac{1}{\cos^2 x}]$
$\:\:\:=\large\frac{(1-\cos x)(1+\cos^2 x+\cos x)}{3x^2.\;\cos^2 x}$ $[since \:\: a^3-b^3=(a-b)(a^2+ab+b^2)]$
$\:\:\:=\large\frac{(2\sin^2 \large\frac{x}{2})(1+\cos^2 x+\cos x)}{3x^2.\;\cos^2 x}$ $[since \:\:1-\cos x=2sin^2 \large\frac{x}{2} ]$
$\Rightarrow\:\lim\limits_{x\to 0}\large\frac{\tan x-\sin x}{x^3}$=$\lim\limits_{x\to 0}\large\frac{(2\sin^2 \large\frac{x}{2})(1+\cos^2 x+\cos x)}{3x^2.\;\cos^2 x}$
$\Rightarrow$$ \lim\limits_{x\to 0}\large\frac{2}{3}.\frac{sin^2 x/2}{x^2}.\frac{1+cos^2 x+cos x}{cos^2 x}$
Applying limits
$=\large\frac{2}{3}.\frac{1}{4}.\frac{3}{1}=\frac{1}{2}$
Hence (c) is the correct answer.
answered Dec 24, 2013 by sreemathi.v
edited Jul 19, 2014 by rvidyagovindarajan_1
 

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