Question Number 126427 by Study last updated on 20/Dec/20
$${li}\underset{{x}\rightarrow−\mathrm{3}} {{m}}\frac{\mid{x}+\mathrm{3}\mid}{{x}+\mathrm{3}}\:\:\:\:\:\:{show}\:{right}\:{and}\:{left}\:{side} \\ $$$${limit}?? \\ $$
Answered by liberty last updated on 20/Dec/20
![lim_(x→−3^− ) [ ((∣x+3∣)/(x+3)) ] = lim_(x→−3^− ) ((−(x+3))/(x+3)) = −1 lim_(x→−3^+ ) [ ((∣x+3∣)/(x+3)) ] = lim_(x→−3^+ ) ((x+3)/(x+3)) = 1](https://www.tinkutara.com/question/Q126446.png)
$$\underset{{x}\rightarrow−\mathrm{3}^{−} } {\mathrm{lim}}\left[\:\frac{\mid{x}+\mathrm{3}\mid}{{x}+\mathrm{3}}\:\right]\:=\:\underset{{x}\rightarrow−\mathrm{3}^{−} } {\mathrm{lim}}\:\frac{−\left({x}+\mathrm{3}\right)}{{x}+\mathrm{3}}\:=\:−\mathrm{1} \\ $$$$\underset{{x}\rightarrow−\mathrm{3}^{+} \:} {\mathrm{lim}}\:\left[\:\frac{\mid{x}+\mathrm{3}\mid}{{x}+\mathrm{3}}\:\right]\:=\:\underset{{x}\rightarrow−\mathrm{3}^{+} } {\mathrm{lim}}\:\frac{{x}+\mathrm{3}}{{x}+\mathrm{3}}\:=\:\mathrm{1} \\ $$$$ \\ $$
Commented by liberty last updated on 20/Dec/20
Commented by Study last updated on 21/Dec/20
$${i}\:{don}'{t}\:{know} \\ $$
Commented by liberty last updated on 22/Dec/20
$${good}\:! \\ $$