Menu Close

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
         b^★ o^★ b^★ h^□ a^□ n^□ s^♠   lim_(x→0)  ((tan 8x−sin 4x.cos 4x)/(x.sin 4x.tan 8x)) ?
$$\:\:\:\:\:\:\:\:\:\overset{\bigstar} {{b}}\overset{\bigstar} {{o}}\overset{\bigstar} {{b}}\overset{\Box} {{h}}\overset{\Box} {{a}}\overset{\Box} {{n}}\overset{\spadesuit} {{s}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{8}{x}−\mathrm{sin}\:\mathrm{4}{x}.\mathrm{cos}\:\mathrm{4}{x}}{{x}.\mathrm{sin}\:\mathrm{4}{x}.\mathrm{tan}\:\mathrm{8}{x}}\:? \\ $$
Answered by john santu last updated on 20/Aug/20
     •J_⊸ S_⊸ •   lim_(x→0)  ((tan 8x−(1/2)sin 8x)/(x sin 4x tan 8x)) =   lim_(x→0)  ((tan 8x(1−(1/2)cos 8x))/(x sin 4x tan 8x)) =  lim_(x→0) ((1−(1/2)cos 8x)/(x sin 4x )) = ∞ (do not exist )
$$\:\:\:\:\:\bullet\underset{\multimap} {{J}}\underset{\multimap} {{S}}\bullet \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{8}{x}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{8}{x}}{{x}\:\mathrm{sin}\:\mathrm{4}{x}\:\mathrm{tan}\:\mathrm{8}{x}}\:=\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{8}{x}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{8}{x}\right)}{{x}\:\mathrm{sin}\:\mathrm{4}{x}\:\mathrm{tan}\:\mathrm{8}{x}}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{8}{x}}{{x}\:\mathrm{sin}\:\mathrm{4}{x}\:}\:=\:\infty\:\left({do}\:{not}\:{exist}\:\right) \\ $$$$ \\ $$
Answered by Dwaipayan Shikari last updated on 20/Aug/20
lim_(x→0) ((tan8x−(1/2)sin8x)/(xsin^2 4x))cos4x=((8x−4x)/(x(4x)^2 ))→∞    (tanx→x ,sinx→x)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{tan}\mathrm{8}{x}−\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{8}{x}}{{xsin}^{\mathrm{2}} \mathrm{4}{x}}{cos}\mathrm{4}{x}=\frac{\mathrm{8}{x}−\mathrm{4}{x}}{{x}\left(\mathrm{4}{x}\right)^{\mathrm{2}} }\rightarrow\infty\:\:\:\:\left({tanx}\rightarrow{x}\:,{sinx}\rightarrow{x}\right) \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *