The given limit $ = \mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {2\sin \frac{x}{2}\cos \frac{x}{2} – 2\sin \frac{x}{2}} \right)}^2} + {{\left( {2{{\sin }^2}\frac{x}{2}} \right)}^3}}}{{2{{\sin }^2}x\cos x – 8\cos x{{\sin }^2}\frac{x}{2} – \frac{4}{3}{{\sin }^4}x}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{4{{\sin }^2}\frac{x}{2}{{\left( {\cos \frac{x}{2} – 1} \right)}^2} + 8{{\sin }^6}\frac{x}{2}}}{{8{{\sin }^2}\frac{x}{2}{{\cos }^2}\frac{x}{2}\cos x – 8\cos x{{\sin }^2}\frac{x}{2} – \frac{4}{3}.16{{\sin }^4}\frac{x}{2}{{\cos }^4}\frac{x}{2}}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{4{{\sin }^2}\frac{x}{2}.4{{\sin }^4}\frac{x}{4} + 8{{\sin }^6}\frac{x}{2}}}{{8{{\sin }^2}\frac{x}{2}\cos x\left( {{{\cos }^2}\frac{x}{2} – 1} \right) – \frac{{64}}{3}{{\sin }^4}\frac{x}{2}{{\cos }^4}\frac{x}{2}}}$ Continue reading $\mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {\sin x – 2\sin \frac{x}{2}} \right)}^2} + {{(1 – \cos x)}^3}}}{{\sin x\sin 2x – 8\cos x{{\sin }^2}\frac{x}{2} – \frac{4}{3}{{\sin }^4}x}} = ?$ →

The given limit,

$ = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {1 + \sin x + \frac{{{{\sin }^2}x}}{{2!}} + \frac{{{{\sin }^3}x}}{{3!}} + ……….} \right) – \left( {1 – \sin x + \frac{{{{\sin }^2}x}}{{2!}} – \frac{{{{\sin }^3}x}}{{3!}} + ……….} \right) – 2\left( {x + \frac{{{x^3}}}{3} + \frac{2}{{15}}{x^5} + ……….} \right)}}{{\left( {x + \frac{{{x^3}}}{3} + \frac{2}{{15}}{x^5} + ……….} \right) – x}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{2\left( {\sin x + \frac{{{{\sin }^3}x}}{{3!}} + \frac{{{{\sin }^5}x}}{{5!}} + ……….} \right) – 2\left( {x + \frac{{{x^3}}}{3} + \frac{2}{{15}}{x^5} + ……….} \right)}}{{\frac{{{x^3}}}{3} + \frac{2}{{15}}{x^5} + ……….}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{2\left( {\sin x – x + \frac{{{{\sin }^3}x}}{{3!}} – \frac{{{x^3}}}{3} + \frac{{{{\sin }^5}x}}{{5!}} – \frac{2}{{15}}{x^5} + ……….} \right)}}{{\frac{{{x^3}}}{3} + \frac{2}{{15}}{x^5} + ……….}}$ Continue reading $\mathop {\lim }\limits_{x \to 0} \frac{{{e^{\sin x}} – {e^{ – \sin x}} – 2\tan x}}{{\tan x – x}} = ?$ →

The given limit,

$ = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{{{(2\cos 2x\cos x + 2\sin x\cos x)}^2}}}{{{{( – 2\cos 2x\sin x + 2\cos 2x)}^3}}}$

$ = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{{{\cos }^2}x{{(\cos 2x + \sin x)}^2}}}{{2{{\cos }^3}2x{{( – \sin x + 1)}^3}}}$

$\mathop { = \lim }\limits_{x \to \frac{\pi }{2}} \frac{{(1 + \sin x)(1 – \sin x){{(\cos 2x + \sin x)}^2}}}{{2{{\cos }^3}2x{{( – \sin x + 1)}^3}}}$ Continue reading Evaluate,

$\mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{{{(\cos x + \cos 3x + \sin 2x)}^2}}}{{{{(\sin x – \sin 3x + 2\cos 2x)}^3}}}$ →

The given limit = ${e^{\mathop {\lim }\limits_{x \to 0} \frac{{\tan x – \sin x}}{{{{\sin }^3}x}}}}$

$ = {e^{\mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{\cos x}} – 1}}{{{{\sin }^2}x}}}}$

$ = {e^{\mathop {\lim }\limits_{x \to 0} \frac{{1 – \cos x}}{{\cos x.{{\sin }^2}x}}}}$

$ = {e^{\mathop {\lim }\limits_{x \to 0} \frac{{1 – \cos x}}{{\cos x.(1 – {{\cos }^2}x)}}}}$

$ = {e^{\mathop {\lim }\limits_{x \to 0} \frac{1}{{\cos x.(1 + \cos x)}}}}$

$ = {e^{1/2}} = \sqrt e $