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Question Number 114405 by bemath last updated on 19/Sep/20
    lim_(x→0)  ((x tan x)/( (√3) cos x−sin^2 x−(√3))) ?
$$\:\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\:\mathrm{tan}\:{x}}{\:\sqrt{\mathrm{3}}\:\mathrm{cos}\:{x}−\mathrm{sin}\:^{\mathrm{2}} {x}−\sqrt{\mathrm{3}}}\:? \\ $$
Answered by bobhans last updated on 19/Sep/20
 lim_(x→0)  ((x tan x)/( (√3)(cos x−1)−sin^2 x)) =  lim_(x→0)  (x^2 /( (√3)(1−(x^2 /2)−1)−x^2 )) =  lim_(x→0)  (x^2 /(x^2 (−((√3)/2)−1))) = ((−2)/( (√3)+2)) = ((−2(2−(√3)))/(4−3))                                         = 2(√3)−2
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\:\mathrm{tan}\:{x}}{\:\sqrt{\mathrm{3}}\left(\mathrm{cos}\:{x}−\mathrm{1}\right)−\mathrm{sin}\:^{\mathrm{2}} {x}}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{3}}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{1}\right)−{x}^{\mathrm{2}} }\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} \left(−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\mathrm{1}\right)}\:=\:\frac{−\mathrm{2}}{\:\sqrt{\mathrm{3}}+\mathrm{2}}\:=\:\frac{−\mathrm{2}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)}{\mathrm{4}−\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{2} \\ $$
Commented by bemath last updated on 19/Sep/20
gave kudos ✓≎
$${gave}\:{kudos}\:\checkmark\Bumpeq \\ $$
Answered by Dwaipayan Shikari last updated on 19/Sep/20
lim_(x→0) ((xtanx)/( (√3)(cosx−1)−sin^2 x))=(x^2 /(−2(√3)sin^2 (x/2)−4sin^2 (x/4)cos^2 (x/2)))  =(x^2 /(((−(√3))/2)x^2 −x^2 ))=−(2/( (√3)+2))=2((√3)−2)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{xtanx}}{\:\sqrt{\mathrm{3}}\left({cosx}−\mathrm{1}\right)−{sin}^{\mathrm{2}} {x}}=\frac{{x}^{\mathrm{2}} }{−\mathrm{2}\sqrt{\mathrm{3}}{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}−\mathrm{4}{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{4}}{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}} \\ $$$$=\frac{{x}^{\mathrm{2}} }{\frac{−\sqrt{\mathrm{3}}}{\mathrm{2}}{x}^{\mathrm{2}} −{x}^{\mathrm{2}} }=−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}+\mathrm{2}}=\mathrm{2}\left(\sqrt{\mathrm{3}}−\mathrm{2}\right) \\ $$

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