Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

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

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{−\mathrm{10cosxsinx}+\mathrm{2sinx}}{−\mathrm{sinx}+\mathrm{3sin3x}}=\frac{−\mathrm{10cosx}+\mathrm{2}}{−\mathrm{1}+\mathrm{3}\frac{\mathrm{sin3x}}{\mathrm{sinx}}}=\frac{−\mathrm{10}+\mathrm{2}}{−\mathrm{1}+\mathrm{3}.\mathrm{3}}=−\mathrm{1} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com