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 68885 by aliesam last updated on 16/Sep/19

if f(x)=((∣x∣)/x) g(x)=x^2 −1 find lim_(x→1) f(g(x))

$${if}\: \\ $$$${f}\left({x}\right)=\frac{\mid{x}\mid}{{x}} \\ $$$${g}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{1} \\ $$$$ \\ $$$${find}\: \\ $$$$ \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {{lim}}\:\:{f}\left({g}\left({x}\right)\right) \\ $$

Commented by kaivan.ahmadi last updated on 16/Sep/19

f(g(x))=f(x^2 −1)=((∣x^2 −1∣)/(x^2 −1)) lim_(x→1^+ ) ((∣x^2 −1∣)/(x^2 −1))=lim_(x→1^+ ) ((x^2 −1)/(x^2 −1))=1 lim_(x→1^− ) ((∣x^2 −1∣)/(x^2 −1))=lim_(x→1^− ) ((−(x^2 −1))/(x^2 −1))=−1 ⇒lim_(x→1) f(g(x)) is not exist.

$${f}\left({g}\left({x}\right)\right)={f}\left({x}^{\mathrm{2}} −\mathrm{1}\right)=\frac{\mid{x}^{\mathrm{2}} −\mathrm{1}\mid}{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$${lim}_{{x}\rightarrow\mathrm{1}^{+} } \frac{\mid{x}^{\mathrm{2}} −\mathrm{1}\mid}{{x}^{\mathrm{2}} −\mathrm{1}}={lim}_{{x}\rightarrow\mathrm{1}^{+} } \frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}=\mathrm{1} \\ $$$${lim}_{{x}\rightarrow\mathrm{1}^{−} } \frac{\mid{x}^{\mathrm{2}} −\mathrm{1}\mid}{{x}^{\mathrm{2}} −\mathrm{1}}={lim}_{{x}\rightarrow\mathrm{1}^{−} } \frac{−\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}=−\mathrm{1} \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{1}} \:{f}\left({g}\left({x}\right)\right)\:{is}\:{not}\:{exist}. \\ $$

Commented by aliesam last updated on 16/Sep/19

if lim_(x→c) g(x) then lim_(x→c) f(g(x)) =f(lim_(x→c) g(x)) =f(b)

$${if} \\ $$$$\underset{{x}\rightarrow{c}} {{lim}}\:{g}\left({x}\right) \\ $$$${then} \\ $$$$\underset{{x}\rightarrow{c}} {{lim}}\:{f}\left({g}\left({x}\right)\right) \\ $$$$={f}\left(\underset{{x}\rightarrow{c}} {{lim}}\:{g}\left({x}\right)\right) \\ $$$$={f}\left({b}\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com