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))](https://www.tinkutara.com/question/Q68885.png)
$${if}\: \\ $$$${f}\left({x}\right)=\frac{\mid{x}\mid}{{x}} \\ $$$${g}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{1} \\ $$$$ \\ $$$${find}\: \\ $$$$ \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {{lim}}\:\:{f}\left({g}\left({x}\right)\right) \\ $$
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.](https://www.tinkutara.com/question/Q68890.png)
$${f}\left({g}\left({x}\right)\right)={f}\left({x}^{\mathrm{2}} −\mathrm{1}\right)=\frac{\mid{x}^{\mathrm{2}} −\mathrm{1}\mid}{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$${lim}_{{x}\rightarrow\mathrm{1}^{+} } \frac{\mid{x}^{\mathrm{2}} −\mathrm{1}\mid}{{x}^{\mathrm{2}} −\mathrm{1}}={lim}_{{x}\rightarrow\mathrm{1}^{+} } \frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{1}}=\mathrm{1} \\ $$$${lim}_{{x}\rightarrow\mathrm{1}^{−} } \frac{\mid{x}^{\mathrm{2}} −\mathrm{1}\mid}{{x}^{\mathrm{2}} −\mathrm{1}}={lim}_{{x}\rightarrow\mathrm{1}^{−} } \frac{−\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}=−\mathrm{1} \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{1}} \:{f}\left({g}\left({x}\right)\right)\:{is}\:{not}\:{exist}. \\ $$
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)](https://www.tinkutara.com/question/Q68897.png)
$${if} \\ $$$$\underset{{x}\rightarrow{c}} {{lim}}\:{g}\left({x}\right) \\ $$$${then} \\ $$$$\underset{{x}\rightarrow{c}} {{lim}}\:{f}\left({g}\left({x}\right)\right) \\ $$$$={f}\left(\underset{{x}\rightarrow{c}} {{lim}}\:{g}\left({x}\right)\right) \\ $$$$={f}\left({b}\right) \\ $$