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 161586 by blackmamba last updated on 20/Dec/21

lim_(x→2^− ) ((2−2cos (√(x−2)))/((x−2)^2 )) =?

$$\:\:\underset{{x}\rightarrow\mathrm{2}^{−} } {\mathrm{lim}}\:\frac{\mathrm{2}−\mathrm{2cos}\:\sqrt{{x}−\mathrm{2}}}{\left({x}−\mathrm{2}\right)^{\mathrm{2}} }\:=? \\ $$

Commented by cortano last updated on 20/Dec/21

Commented by mathmax by abdo last updated on 20/Dec/21

sir cortano you take another limit loook the Q...

$$\mathrm{sir}\:\mathrm{cortano}\:\mathrm{you}\:\mathrm{take}\:\mathrm{another}\:\mathrm{limit}\:\mathrm{loook}\:\mathrm{the}\:\mathrm{Q}... \\ $$

Answered by TheSupreme last updated on 20/Dec/21

t=x−2 lim ((2−2cos((√t)))/t^2 )=2((1−cos((√t)))/t^2 )=2((1−(1−(t/2)))/t^2 )=∞

$${t}={x}−\mathrm{2} \\ $$$${lim}\:\frac{\mathrm{2}−\mathrm{2}{cos}\left(\sqrt{{t}}\right)}{{t}^{\mathrm{2}} }=\mathrm{2}\frac{\mathrm{1}−{cos}\left(\sqrt{{t}}\right)}{{t}^{\mathrm{2}} }=\mathrm{2}\frac{\mathrm{1}−\left(\mathrm{1}−\frac{{t}}{\mathrm{2}}\right)}{{t}^{\mathrm{2}} }=\infty \\ $$

Commented by cortano last updated on 20/Dec/21

the question lim_(x→2^− ) ((x−2cos (√(2−x)))/((2−x)^2 )) (√(2−x)) defined on 2−x≥0 ⇒x≤2

$${the}\:{question} \\ $$$$\:\underset{{x}\rightarrow\mathrm{2}^{−} } {\mathrm{lim}}\:\frac{{x}−\mathrm{2cos}\:\sqrt{\mathrm{2}−{x}}}{\left(\mathrm{2}−{x}\right)^{\mathrm{2}} } \\ $$$$\:\sqrt{\mathrm{2}−{x}}\:{defined}\:{on}\:\mathrm{2}−{x}\geqslant\mathrm{0} \\ $$$$\:\Rightarrow{x}\leqslant\mathrm{2}\: \\ $$

Commented by cortano last updated on 20/Dec/21

why not valid?

$${why}\:{not}\:{valid}? \\ $$

Commented by cortano last updated on 20/Dec/21

Answer −(1/(12))

$${Answer}\:−\frac{\mathrm{1}}{\mathrm{12}} \\ $$

Commented by mr W last updated on 20/Dec/21

−(1/(12)) is wrong. lim_(x→2^+ ) (...)=+∞ x<2 is not valid.

$$−\frac{\mathrm{1}}{\mathrm{12}}\:{is}\:{wrong}. \\ $$$$\underset{{x}\rightarrow\mathrm{2}^{+} } {\mathrm{lim}}\:\left(...\right)=+\infty \\ $$$${x}<\mathrm{2}\:{is}\:{not}\:{valid}. \\ $$

Commented by mnjuly1970 last updated on 20/Dec/21

D_( f) = (∞ ,2]

$$\:\:\:\mathrm{D}_{\:{f}} \:\:=\:\left(\infty\:,\mathrm{2}\right] \\ $$

Commented by mr W last updated on 20/Dec/21

cortano sir: you changed the question by yourself. the question is ((2−2 cos (√(x−2)))/((x−2)^2 )), not ((x−2 cos (√(2−x)))/((2−x)^2 )) !

$${cortano}\:{sir}: \\ $$$${you}\:{changed}\:{the}\:{question}\:{by}\:{yourself}. \\ $$$${the}\:{question}\:{is} \\ $$$$\frac{\mathrm{2}−\mathrm{2}\:\mathrm{cos}\:\sqrt{{x}−\mathrm{2}}}{\left({x}−\mathrm{2}\right)^{\mathrm{2}} },\:{not}\:\frac{{x}−\mathrm{2}\:\mathrm{cos}\:\sqrt{\mathrm{2}−{x}}}{\left(\mathrm{2}−{x}\right)^{\mathrm{2}} }\:! \\ $$

Answered by mathmax by abdo last updated on 20/Dec/21

f(x)=((2−2cos(√(x−2)))/((x−2)^2 )) we do the changement (√(x−2))=t ⇒ f(x)=((2−2cost)/t^4 ) (x→2 ⇒t→0) cost∼1−(t^2 /2) ⇒f(x)∼((2−2+t^2 )/t^4 )=(1/t^2 ) →+∞

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{2}−\mathrm{2cos}\sqrt{\mathrm{x}−\mathrm{2}}}{\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{2}} }\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\sqrt{\mathrm{x}−\mathrm{2}}=\mathrm{t}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{2}−\mathrm{2cost}}{\mathrm{t}^{\mathrm{4}} }\:\:\:\:\:\:\left(\mathrm{x}\rightarrow\mathrm{2}\:\Rightarrow\mathrm{t}\rightarrow\mathrm{0}\right) \\ $$$$\mathrm{cost}\sim\mathrm{1}−\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\sim\frac{\mathrm{2}−\mathrm{2}+\mathrm{t}^{\mathrm{2}} }{\mathrm{t}^{\mathrm{4}} }=\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }\:\rightarrow+\infty \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com