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lim-x-0-5cos-2-x-2cos-x-3-cos-x-cos-3x-




Question Number 110132 by bemath last updated on 27/Aug/20
 lim_(x→0) ((5cos^2 x−2cos x−3)/(cos x−cos 3x)) ?
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{5cos}\:^{\mathrm{2}} {x}−\mathrm{2cos}\:{x}−\mathrm{3}}{\mathrm{cos}\:{x}−\mathrm{cos}\:\mathrm{3}{x}}\:? \\ $$
Commented by PRITHWISH SEN 2 last updated on 27/Aug/20
form (0/0) use L′Hopital
$$\mathrm{form}\:\frac{\mathrm{0}}{\mathrm{0}}\:\mathrm{use}\:\mathrm{L}'\mathrm{Hopital} \\ $$
Commented by bemath last updated on 27/Aug/20
Commented by bemath last updated on 27/Aug/20
yes...cooll
$${yes}…{cooll} \\ $$
Commented by Dwaipayan Shikari last updated on 27/Aug/20
lim_(x→0) ((−10cosxsinx+2sinx)/(−sinx+3sin3x))=((−10cosx+2)/(−1+3((sin3x)/(sinx))))=((−10+2)/(−1+3.3))=−1
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{10cosxsinx}+\mathrm{2sinx}}{−\mathrm{sinx}+\mathrm{3sin3x}}=\frac{−\mathrm{10cosx}+\mathrm{2}}{−\mathrm{1}+\mathrm{3}\frac{\mathrm{sin3x}}{\mathrm{sinx}}}=\frac{−\mathrm{10}+\mathrm{2}}{−\mathrm{1}+\mathrm{3}.\mathrm{3}}=−\mathrm{1} \\ $$

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