Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 142333 by mohammad17 last updated on 30/May/21

Commented by mohammad17 last updated on 30/May/21

please sir help me

$${please}\:{sir}\:{help}\:{me} \\ $$

Answered by Dwaipayan Shikari last updated on 30/May/21

lim_(x→0) (((x+4)^(3/2) −8)/x)=((8((1+(x/4))^(3/2) −1))/x)=8((1+(x/4).(3/2)−1)/x)=3

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({x}+\mathrm{4}\right)^{\mathrm{3}/\mathrm{2}} −\mathrm{8}}{{x}}=\frac{\mathrm{8}\left(\left(\mathrm{1}+\frac{{x}}{\mathrm{4}}\right)^{\mathrm{3}/\mathrm{2}} −\mathrm{1}\right)}{{x}}=\mathrm{8}\frac{\mathrm{1}+\frac{{x}}{\mathrm{4}}.\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{1}}{{x}}=\mathrm{3} \\ $$

Commented by mathmax by abdo last updated on 30/May/21

let f(x)=(((x+4)^(3/2) −8)/x) ⇒f(x)=((4^(3/2) (1+(x/4))^(3/2) −8)/x) =((8{(1+(x/4))^(3/2) −1})/x)∼((8(1+((3x)/8)−1))/x)=3 ⇒lim_(x→0) f(x)=3

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\left(\mathrm{x}+\mathrm{4}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\mathrm{8}}{\mathrm{x}}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{4}^{\frac{\mathrm{3}}{\mathrm{2}}} \left(\mathrm{1}+\frac{\mathrm{x}}{\mathrm{4}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\mathrm{8}}{\mathrm{x}} \\ $$$$=\frac{\mathrm{8}\left\{\left(\mathrm{1}+\frac{\mathrm{x}}{\mathrm{4}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\mathrm{1}\right\}}{\mathrm{x}}\sim\frac{\mathrm{8}\left(\mathrm{1}+\frac{\mathrm{3x}}{\mathrm{8}}−\mathrm{1}\right)}{\mathrm{x}}=\mathrm{3}\:\Rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right)=\mathrm{3} \\ $$$$ \\ $$

Answered by bramlexs22 last updated on 30/May/21

f(x)=lim_(x→0) (((√((x+4)^3 ))−8)/x) f(x)=lim_(x→0) (((x+4)^3 −64)/x).lim_(x→0) (1/( (√((x+4)^3 ))+8)) = lim_(x→0) ((x(x^2 +12x+48))/x). (1/(16)) = ((48)/(16)) = 3

$${f}\left({x}\right)=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\left({x}+\mathrm{4}\right)^{\mathrm{3}} }−\mathrm{8}}{{x}} \\ $$$${f}\left({x}\right)=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({x}+\mathrm{4}\right)^{\mathrm{3}} −\mathrm{64}}{{x}}.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt{\left({x}+\mathrm{4}\right)^{\mathrm{3}} }+\mathrm{8}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\left({x}^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{48}\right)}{{x}}.\:\frac{\mathrm{1}}{\mathrm{16}} \\ $$$$=\:\frac{\mathrm{48}}{\mathrm{16}}\:=\:\mathrm{3}\: \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com