Question Number 158760 by mathlove last updated on 08/Nov/21
![f(x)=[sgn(x^2 −1)+sgn(sin πx)] faind lim_(x→1) f(x)=?](Q158760.png)
$${f}\left({x}\right)=\left[{sgn}\left({x}^{\mathrm{2}} −\mathrm{1}\right)+{sgn}\left(\mathrm{sin}\:\pi{x}\right)\right] \\ $$$${faind}\:\:\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}{f}\left({x}\right)=? \\ $$
Commented by mathlove last updated on 08/Nov/21
$${mister}\:{W}\:{snswer}\:{the}\:{Q} \\ $$$${andAjfur} \\ $$
Commented by mathlove last updated on 08/Nov/21
$${ajfuor} \\ $$
Answered by mr W last updated on 08/Nov/21
![f(1)=[0+0]=0 lim_(x→1^− ) f(x)=[−1+1]=0 lim_(x→1^+ ) f(x)=[1−1]=0 ⇒lim_(x→1) f(x)=0](Q158778.png)
$${f}\left(\mathrm{1}\right)=\left[\mathrm{0}+\mathrm{0}\right]=\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{1}^{−} } {\mathrm{lim}}{f}\left({x}\right)=\left[−\mathrm{1}+\mathrm{1}\right]=\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{1}^{+} } {\mathrm{lim}}{f}\left({x}\right)=\left[\mathrm{1}−\mathrm{1}\right]=\mathrm{0} \\ $$$$\Rightarrow\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}{f}\left({x}\right)=\mathrm{0} \\ $$
Commented by mathlove last updated on 09/Nov/21
$${thanks}\:{sir} \\ $$