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 115195 by bemath last updated on 24/Sep/20

lim_(x→(π/8))  ((cot 4x−cos 4x)/((π−8x)^3 )) ?

$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{8}}} {\mathrm{lim}}\:\frac{\mathrm{cot}\:\mathrm{4}{x}−\mathrm{cos}\:\mathrm{4}{x}}{\left(\pi−\mathrm{8}{x}\right)^{\mathrm{3}} }\:?\: \\ $$

Answered by bobhans last updated on 24/Sep/20

let x = (π/8)+p ; p→0  lim_(p→0)  ((cot (4p+(π/2))−cos (4p+(π/2)))/((−8p)^3 )) =  lim_(p→0)  ((−tan 4p+sin 4p)/(−512p^3 )) =   lim_(p→0)  ((tan 4p−sin 4p)/(512p^3 )) = lim_(p→0)  ((tan 4p(1−cos 4p))/(512p^3 ))  = ((4×8)/(64×8)) = (1/(16))

$${let}\:{x}\:=\:\frac{\pi}{\mathrm{8}}+{p}\:;\:{p}\rightarrow\mathrm{0} \\ $$$$\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cot}\:\left(\mathrm{4}{p}+\frac{\pi}{\mathrm{2}}\right)−\mathrm{cos}\:\left(\mathrm{4}{p}+\frac{\pi}{\mathrm{2}}\right)}{\left(−\mathrm{8}{p}\right)^{\mathrm{3}} }\:= \\ $$$$\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{tan}\:\mathrm{4}{p}+\mathrm{sin}\:\mathrm{4}{p}}{−\mathrm{512}{p}^{\mathrm{3}} }\:=\: \\ $$$$\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{4}{p}−\mathrm{sin}\:\mathrm{4}{p}}{\mathrm{512}{p}^{\mathrm{3}} }\:=\:\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\mathrm{4}{p}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{4}{p}\right)}{\mathrm{512}{p}^{\mathrm{3}} } \\ $$$$=\:\frac{\mathrm{4}×\mathrm{8}}{\mathrm{64}×\mathrm{8}}\:=\:\frac{\mathrm{1}}{\mathrm{16}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com