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 71317 by aliesam last updated on 13/Oct/19

lim_(x→0) ((∣2x−2∣−∣2x+2∣)/x)

$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{\mid\mathrm{2}{x}−\mathrm{2}\mid−\mid\mathrm{2}{x}+\mathrm{2}\mid}{{x}} \\ $$

Commented by mathmax by abdo last updated on 13/Oct/19

=lim_(x→0) (((√((2x−2)^2 ))−(√((2x+2)^2 )))/x) =lim_(x→0) ((((√((2x−2)^2 ))−(√((2x+2)^2 )))((√((2x−2)^2 ))+(√((2x+2)^2 ))))/(x{(√((2x−2)^2 ))+(√((2x+2)^2 ))})) =lim_(x→0) (((2x−2)^2 −(2x+2)^2 )/(x{(√((2x−2)^2 ))+(√((2x+2)^2 ))})) =lim_(x→0) ((4(x^2 −2x+1−x^2 −2x−1})/(x{(√((2x−2)^2 ))+(√((2x+2)^2 ))})) =lim_(x→0) ((−16)/((√((2x−2)^2 ))+(√((2x+2)^2 )))) =((−16)/4) =−4

$$={lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{\sqrt{\left(\mathrm{2}{x}−\mathrm{2}\right)^{\mathrm{2}} }−\sqrt{\left(\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{2}} }}{{x}} \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{\left(\sqrt{\left(\mathrm{2}{x}−\mathrm{2}\right)^{\mathrm{2}} }−\sqrt{\left(\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{2}} }\right)\left(\sqrt{\left(\mathrm{2}{x}−\mathrm{2}\right)^{\mathrm{2}} }+\sqrt{\left(\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{2}} }\right)}{{x}\left\{\sqrt{\left(\mathrm{2}{x}−\mathrm{2}\right)^{\mathrm{2}} }+\sqrt{\left(\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{2}} }\right\}} \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{\left(\mathrm{2}{x}−\mathrm{2}\right)^{\mathrm{2}} −\left(\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{2}} }{{x}\left\{\sqrt{\left(\mathrm{2}{x}−\mathrm{2}\right)^{\mathrm{2}} }+\sqrt{\left(\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{2}} }\right\}} \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{\mathrm{4}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}−{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{1}\right\}}{{x}\left\{\sqrt{\left(\mathrm{2}{x}−\mathrm{2}\right)^{\mathrm{2}} }+\sqrt{\left(\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{2}} }\right\}} \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\:\frac{−\mathrm{16}}{\sqrt{\left(\mathrm{2}{x}−\mathrm{2}\right)^{\mathrm{2}} }+\sqrt{\left(\mathrm{2}{x}+\mathrm{2}\right)^{\mathrm{2}} }}\:=\frac{−\mathrm{16}}{\mathrm{4}}\:=−\mathrm{4} \\ $$

Commented by aliesam last updated on 13/Oct/19

god bless you sir

$${god}\:{bless}\:{you}\:{sir} \\ $$

Commented by mathmax by abdo last updated on 13/Oct/19

thank you sir.

$${thank}\:{you}\:{sir}. \\ $$

Answered by ajfour last updated on 13/Oct/19

x=h→0^+ L=lim_(h→0) ((2−2h−2h−2)/h)=−4 x=−h→0^− L=lim_(h→0) ((2+2h+2h−2)/(−h))=−4 hence limit L does exist and=−4.

$${x}={h}\rightarrow\mathrm{0}^{+} \\ $$$${L}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}−\mathrm{2}{h}−\mathrm{2}{h}−\mathrm{2}}{{h}}=−\mathrm{4} \\ $$$${x}=−{h}\rightarrow\mathrm{0}^{−} \\ $$$${L}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}+\mathrm{2}{h}+\mathrm{2}{h}−\mathrm{2}}{−{h}}=−\mathrm{4} \\ $$$${hence}\:{limit}\:{L}\:{does}\:{exist}\:{and}=−\mathrm{4}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com