Question Number 38111 by maxmathsup by imad last updated on 21/Jun/18
![find lim_(x→0) ((e^x −[x])/x)](https://www.tinkutara.com/question/Q38111.png)
$${find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\frac{{e}^{{x}} \:−\left[{x}\right]}{{x}} \\ $$
Commented by math khazana by abdo last updated on 22/Jun/18
![theQ is find lim_(x→0) ((e^x −1 −[x])/x)](https://www.tinkutara.com/question/Q38129.png)
$${theQ}\:{is}\:{find}\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\frac{{e}^{{x}} \:−\mathrm{1}\:−\left[{x}\right]}{{x}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jun/18
![[x]=0 when x→0+ lim_(x→0+) ((e^x −[x])/x) =lim_(x→0+) (e^x /x)=∞ [x]→−1 when x→0− lim_(x→0−) ((e^x −[x])/x) =lim_(x→0−) ((e^x −1)/x)=1 if lim_(x→0) ((e^x −1−[x])/x) lim_(x→0−) ((e^x −1+1)/x) { [x]→−1 when x→0− =lim_(x→0−) (e^x /x)=∞ li_(x→0−) lim_(x→0+) ((e^x −1−[x])/x) =lim_(x→0+) ((e^x −1−0)/x) =1](https://www.tinkutara.com/question/Q38176.png)
$$\left[{x}\right]=\mathrm{0}\:\:{when}\:{x}\rightarrow\mathrm{0}+ \\ $$$$\underset{{x}\rightarrow\mathrm{0}+} {\mathrm{lim}}\frac{{e}^{{x}} −\left[{x}\right]}{{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}+} {\mathrm{lim}}\frac{{e}^{{x}} }{{x}}=\infty \\ $$$$\left[{x}\right]\rightarrow−\mathrm{1}\:{when}\:{x}\rightarrow\mathrm{0}− \\ $$$$\underset{{x}\rightarrow\mathrm{0}−} {\mathrm{lim}}\frac{{e}^{{x}} −\left[{x}\right]}{{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}−} {\mathrm{lim}}\frac{{e}^{{x}} −\mathrm{1}}{{x}}=\mathrm{1} \\ $$$${if} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{x}} −\mathrm{1}−\left[{x}\right]}{{x}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}−} {\mathrm{lim}}\frac{{e}^{{x}} −\mathrm{1}+\mathrm{1}}{{x}}\:\:\left\{\:\left[{x}\right]\rightarrow−\mathrm{1}\:\:\:{when}\:{x}\rightarrow\mathrm{0}−\right. \\ $$$$=\underset{{x}\rightarrow\mathrm{0}−} {\mathrm{lim}}\frac{{e}^{{x}} }{{x}}=\infty \\ $$$$\underset{{x}\rightarrow\mathrm{0}−} {\mathrm{li}} \\ $$$$ \\ $$$$\underset{{x}\rightarrow\mathrm{0}+} {\mathrm{lim}}\frac{{e}^{{x}} −\mathrm{1}−\left[{x}\right]}{{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}+} {\mathrm{lim}}\frac{{e}^{{x}} −\mathrm{1}−\mathrm{0}}{{x}} \\ $$$$=\mathrm{1} \\ $$$$ \\ $$