Question Number 118140 by bemath last updated on 15/Oct/20
![lim_(x→0) (([ x^2 ])/(2x)) =?](https://www.tinkutara.com/question/Q118140.png)
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left[\:{x}^{\mathrm{2}} \:\right]}{\mathrm{2}{x}}\:=?\: \\ $$
Answered by TANMAY PANACEA last updated on 15/Oct/20
![let x=0.9 x^2 =0.81 [0.81]=0 so (([x^2 ])/(2x))=(0/(2×0.9))=0 let x=0.01 x^2 =0.0001 [x^2 ]=0 (([x^2 ])/(2×x))=(0/(2×0.01))=0 let x=−0.01 x^2 =0.0001 (([x^2 ])/(2x))=(0/(−2×0.01))=0 so lim_(x→0) (([x^2 ])/(2x))=0](https://www.tinkutara.com/question/Q118143.png)
$${let}\:{x}=\mathrm{0}.\mathrm{9}\:\:{x}^{\mathrm{2}} =\mathrm{0}.\mathrm{81}\:\:\left[\mathrm{0}.\mathrm{81}\right]=\mathrm{0} \\ $$$${so}\:\frac{\left[{x}^{\mathrm{2}} \right]}{\mathrm{2}{x}}=\frac{\mathrm{0}}{\mathrm{2}×\mathrm{0}.\mathrm{9}}=\mathrm{0} \\ $$$${let}\:{x}=\mathrm{0}.\mathrm{01}\:\:{x}^{\mathrm{2}} =\mathrm{0}.\mathrm{0001}\:\:\left[{x}^{\mathrm{2}} \right]=\mathrm{0} \\ $$$$\frac{\left[{x}^{\mathrm{2}} \right]}{\mathrm{2}×{x}}=\frac{\mathrm{0}}{\mathrm{2}×\mathrm{0}.\mathrm{01}}=\mathrm{0} \\ $$$${let}\:{x}=−\mathrm{0}.\mathrm{01}\:\:\:{x}^{\mathrm{2}} =\mathrm{0}.\mathrm{0001} \\ $$$$\frac{\left[{x}^{\mathrm{2}} \right]}{\mathrm{2}{x}}=\frac{\mathrm{0}}{−\mathrm{2}×\mathrm{0}.\mathrm{01}}=\mathrm{0} \\ $$$$\boldsymbol{{so}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\left[\boldsymbol{{x}}^{\mathrm{2}} \right]}{\mathrm{2}\boldsymbol{{x}}}=\mathrm{0} \\ $$
Commented by bemath last updated on 15/Oct/20
$${thank}\:{you}\:{sir} \\ $$
Commented by TANMAY PANACEA last updated on 15/Oct/20
$${most}\:{welcome}\:{sir} \\ $$
Answered by Bird last updated on 16/Oct/20
![[x^2 ]≤x^2 ≤[x^2 ] +1 ⇒for x>0 (([x^2 ])/(2x))≤(x^2 /(2x))≤(([x^2 ]+1)/(2x)) ⇒(([x^2 ])/(2x))≤(x/2) and lim_(x→0) (x/2)=0 ⇒lim_(x→0^+ ) (([x^2 ])/(2x))=0](https://www.tinkutara.com/question/Q118220.png)
$$\left[{x}^{\mathrm{2}} \right]\leqslant{x}^{\mathrm{2}} \leqslant\left[{x}^{\mathrm{2}} \right]\:+\mathrm{1}\:\Rightarrow{for}\:{x}>\mathrm{0} \\ $$$$\frac{\left[{x}^{\mathrm{2}} \right]}{\mathrm{2}{x}}\leqslant\frac{{x}^{\mathrm{2}} }{\mathrm{2}{x}}\leqslant\frac{\left[{x}^{\mathrm{2}} \right]+\mathrm{1}}{\mathrm{2}{x}}\:\Rightarrow\frac{\left[{x}^{\mathrm{2}} \right]}{\mathrm{2}{x}}\leqslant\frac{{x}}{\mathrm{2}} \\ $$$${and}\:{lim}_{{x}\rightarrow\mathrm{0}} \frac{{x}}{\mathrm{2}}=\mathrm{0}\:\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\frac{\left[{x}^{\mathrm{2}} \right]}{\mathrm{2}{x}}=\mathrm{0} \\ $$$$ \\ $$