# $\lim\limits_{x\to 0}\large\frac{\tan x-\sin x}{x^3}$=
$(a)\;1\qquad(b)\;2\qquad(c)\;\large\frac{1}{2}$$\qquad(d)\;-1 ## 1 Answer Comment A) Toolbox: • L. Hospital,s Rule: \lim\limits_{x\to 0}f(x)=\lim\limits_{x\to 0} f' (x) • \lim\limits_{x\to 0}\large\frac{f(x)}{g(x)}=\lim\limits_{x\to 0}\large\frac{f'(x)}{g'(x)} where g(x)\neq0 \lim\limits_{x\to 0}\large\frac{\tan x-\sin x}{x^3} Let f(x)=\tan x-\sin x and g(x)=x^3 f'(x)=\sec^2 x-\cos x and g'(x)=3x^3 \large \frac{f'(x)}{g'(x)}=\large\frac{\sec^2 x-\cos x}{3x^2} \:\:\:=\large\frac{1-\cos^3 x}{3x^2.\;\cos^2 x} [since \:\sec^2 x=\large\frac{1}{\cos^2 x}] \:\:\:=\large\frac{(1-\cos x)(1+\cos^2 x+\cos x)}{3x^2.\;\cos^2 x} [since \:\: a^3-b^3=(a-b)(a^2+ab+b^2)] \:\:\:=\large\frac{(2\sin^2 \large\frac{x}{2})(1+\cos^2 x+\cos x)}{3x^2.\;\cos^2 x} [since \:\:1-\cos x=2sin^2 \large\frac{x}{2} ] \Rightarrow\:\lim\limits_{x\to 0}\large\frac{\tan x-\sin x}{x^3}=\lim\limits_{x\to 0}\large\frac{(2\sin^2 \large\frac{x}{2})(1+\cos^2 x+\cos x)}{3x^2.\;\cos^2 x} \Rightarrow$$ \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}$
$=\large\frac{2}{3}.\frac{1}{4}.\frac{3}{1}=\frac{1}{2}$