b-o-b-h-a-n-s-lim-x-0-tan-8x-sin-4x-cos-4x-x-sin-4x-tan-8x-

Question Number 108956 by bobhans last updated on 20/Aug/20

\overset{★}{b} \overset{★}{o} \overset{★}{b} \overset{}{h} \overset{}{a} \overset{}{n} \overset{♠}{s}

\lim_{x \to 0} \frac{\tan 8 x - \sin 4 x . \cos 4 x}{x . \sin 4 x . \tan 8 x} ?

Answered by john santu last updated on 20/Aug/20

∙ \underset{⊸}{J} \underset{⊸}{S} ∙

\lim_{x \to 0} \frac{\tan 8 x - \frac{1}{2} \sin 8 x}{x \sin 4 x \tan 8 x} =

\lim_{x \to 0} \frac{\tan 8 x (1 - \frac{1}{2} \cos 8 x)}{x \sin 4 x \tan 8 x} =

\lim_{x \to 0} \frac{1 - \frac{1}{2} \cos 8 x}{x \sin 4 x} = \infty (d o n o t e x i s t)

Answered by Dwaipayan Shikari last updated on 20/Aug/20

\lim_{x \to 0} \frac{t a n 8 x - \frac{1}{2} s i n 8 x}{{x s i n}^{2} 4 x} c o s 4 x = \frac{8 x - 4 x}{x {(4 x)}^{2}} \to \infty (t a n x \to x, s i n x \to x)