Question Number 204360 by esmaeil last updated on 14/Feb/24
$${if} \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left[\frac{\left({a}+\mathrm{6}\right){x}+\mathrm{1}}{{ax}+\mathrm{1}}\right]=\mathrm{2}\rightarrow \\ $$$${find}\:{the}\:{largest}\:{and}\:{smallest}\:{correct} \\ $$$${value}\:{for}\:{a}. \\ $$
Answered by AST last updated on 14/Feb/24
$$=\underset{{x}\rightarrow+\infty} {{lim}}\left[\frac{{a}+\mathrm{6}+\frac{\mathrm{1}}{{x}}}{{a}+\frac{\mathrm{1}}{{x}}}\right]=\frac{{a}+\mathrm{6}}{{a}}=\mathrm{2}\Rightarrow{a}=\mathrm{6} \\ $$
Commented by esmaeil last updated on 14/Feb/24
$${thank}\:\:{you}.\:\:{and}\:{smallest}\:{value}? \\ $$
Commented by esmaeil last updated on 14/Feb/24
$$\mathrm{2}\leqslant\frac{{a}+\mathrm{6}}{{a}}<\mathrm{3}\rightarrow\mathrm{1}\leqslant\frac{\mathrm{6}}{{a}}\langle\mathrm{2}\rightarrow \\ $$$$\mathrm{3}<{a}\leqslant\mathrm{6}\rightarrow\begin{cases}{{a}_{{l}} =\mathrm{6}}\\{{a}_{{s}} =\mathrm{4}\:\:\left({a}\:{is}\:{correct}\right)}\end{cases} \\ $$$$ \\ $$$$ \\ $$
Commented by esmaeil last updated on 14/Feb/24
$$\left[\frac{\mathrm{10}}{\mathrm{4}}\right]=\mathrm{2} \\ $$$$ \\ $$
Commented by AST last updated on 15/Feb/24
$${I}\:{thought}\:\left[\:\right]\:{was}\:{just}\:{a}\:{bracket},{if}\:\left[{x}\right]\:{is}\:{floor}\:{of} \\ $$$${x},\:{then}\:\left[\frac{{a}+\mathrm{6}}{{a}}\right]=\mathrm{2}<\frac{{a}+\mathrm{6}}{{a}}\leqslant\mathrm{3}\Rightarrow\mathrm{3}\leqslant{a}<\mathrm{6} \\ $$$$\Rightarrow{min}\left({a}\right)=\mathrm{3}\:{and}\:{upper}\:{bound}\:{of}\:{a}=\mathrm{6} \\ $$
Answered by mr W last updated on 15/Feb/24
$${let}\:{t}=\frac{\mathrm{1}}{{x}} \\ $$$${x}\rightarrow+\infty\:\Rightarrow{t}\rightarrow\mathrm{0}^{+} \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left[\frac{\left({a}+\mathrm{6}\right){x}+\mathrm{1}}{{ax}+\mathrm{1}}\right] \\ $$$$=\underset{{t}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left[\frac{{a}+\mathrm{6}+{t}}{{a}+{t}}\right]=\mathrm{2} \\ $$$$\Rightarrow\mathrm{2}<\frac{{a}+\mathrm{6}}{{a}}\leqslant\mathrm{3} \\ $$$$\Rightarrow\mathrm{3}\leqslant{a}<\mathrm{6} \\ $$
Commented by AST last updated on 14/Feb/24
$${It}\:{seems}\:\mathrm{6}\:{doesn}'{t}\:{work}. \\ $$$${a}=\mathrm{6}\Rightarrow\underset{{x}\rightarrow+\infty} {{lim}}\left[\frac{\mathrm{12}{x}+\mathrm{1}}{\mathrm{6}{x}+\mathrm{1}}\right]=\underset{{x}\rightarrow+\infty} {{lim}}\left[\frac{\mathrm{2}\left(\mathrm{6}{x}+\mathrm{1}\right)−\mathrm{1}}{\mathrm{6}{x}+\mathrm{1}}\right] \\ $$$$=\underset{{x}\rightarrow+\infty} {{lim}}\left[\mathrm{2}−\frac{\mathrm{1}}{\mathrm{6}{x}+\mathrm{1}}\right]=\mathrm{1},{since}\:\mathrm{1}<\mathrm{2}−\frac{\mathrm{1}}{\mathrm{6}{x}+\mathrm{1}}<\mathrm{2}\:{as} \\ $$$${x}\rightarrow+\infty. \\ $$$${Even}\:{though}\:\underset{{x}\rightarrow+\infty} {{lim}}\mathrm{2}−\frac{\mathrm{1}}{\mathrm{6}{x}+\mathrm{1}}=\mathrm{2},\:{it}\:{approaches}\:\mathrm{2} \\ $$$${from}\:{the}\:{negative}\:{axis}\left(\mathrm{2}^{−} \right),{so}\:{all}\:{values}\:{are}\: \\ $$$${less}\:{than}\:\mathrm{2}. \\ $$
Commented by mr W last updated on 14/Feb/24
$${you}\:{are}\:{right}.\:\mathrm{3}\leqslant{a}<\mathrm{6}. \\ $$
Answered by AST last updated on 14/Feb/24
$$\underset{{x}\rightarrow+\infty} {{lim}}\left[\frac{{ax}+\mathrm{1}+\mathrm{6}{x}}{{ax}+\mathrm{1}}\right]=\underset{{x}\rightarrow+\infty} {{lim}}\left[\frac{\mathrm{6}{x}}{{ax}+\mathrm{1}}+\mathrm{1}\right]=\mathrm{2} \\ $$$${If}\:\mathrm{6}{x}<{ax}+\mathrm{1},{then}\:\frac{\mathrm{6}{x}}{{ax}+\mathrm{1}}<\mathrm{1}\Rightarrow\left[\frac{\mathrm{6}{x}}{{ax}+\mathrm{1}}+\mathrm{1}\right]<\mathrm{2} \\ $$$$\Rightarrow\mathrm{6}{x}\geqslant{ax}+\mathrm{1}\Rightarrow{a}\leqslant\frac{\mathrm{6}{x}−\mathrm{1}}{{x}}=\mathrm{6}−\frac{\mathrm{1}}{{x}}<\mathrm{6}\: \\ $$$${a}<\mathrm{6}\Rightarrow\left[\frac{\mathrm{6}{x}}{{ax}+\mathrm{1}}+\mathrm{1}\right]=\left[\frac{{ax}+\mathrm{1}+\left(\mathrm{6}−{a}\right){x}−\mathrm{1}}{{ax}+\mathrm{1}}+\mathrm{1}\right] \\ $$$$=\left[\mathrm{2}+\frac{\left(\mathrm{6}−{a}\right){x}−\mathrm{1}}{{ax}+\mathrm{1}}\right] \\ $$$$\underset{{x}\rightarrow+\infty} {{lim}}\frac{\left(\mathrm{6}−{a}\right){x}−\mathrm{1}}{{ax}+\mathrm{1}}=\frac{\mathrm{6}−{a}}{{a}}=\frac{\mathrm{6}}{{a}}−\mathrm{1} \\ $$$$\underset{{x}\rightarrow+\infty} {{lim}}\left[\frac{{ax}+\mathrm{1}+\mathrm{6}{x}}{{ax}+\mathrm{1}}\right]=\underset{{x}\rightarrow+\infty} {{lim}}\left[\mathrm{2}+\frac{\left(\mathrm{6}−{a}\right){x}−\mathrm{1}}{{ax}+\mathrm{1}}\right]=\mathrm{2} \\ $$$$\underset{{x}\rightarrow+\infty} {{lim}}\frac{\left(\mathrm{6}−{a}\right){x}−\mathrm{1}}{{ax}+\mathrm{1}}\leqslant\mathrm{1}\Rightarrow\frac{\mathrm{6}}{{a}}−\mathrm{1}\leqslant\mathrm{1}\Rightarrow{a}\geqslant\mathrm{3} \\ $$$$\Rightarrow\mathrm{3}\leqslant{a}<\mathrm{6} \\ $$