Question Number 65013 by Rio Michael last updated on 24/Jul/19
$${why}\:{do}\:{we}\:{divide}\:{each}\:{term}\:{by}\:{n}\:{when}\:{given}\:{the}\:{question} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{3}\:+\mathrm{2}{n}}{\mathrm{1}+{n}}\:? \\ $$
Answered by Tanmay chaudhury last updated on 24/Jul/19
![lim_(n→∞) (((3/n)+2)/((1/n)+1)) =((0+2)/(0+1))=2 [we devide N_r and D_r by highest power n to get 0(zero) because (1/∞)→0]](https://www.tinkutara.com/question/Q65018.png)
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\frac{\mathrm{3}}{{n}}+\mathrm{2}}{\frac{\mathrm{1}}{{n}}+\mathrm{1}} \\ $$$$=\frac{\mathrm{0}+\mathrm{2}}{\mathrm{0}+\mathrm{1}}=\mathrm{2}\:\:\:\left[{we}\:{devide}\:{N}_{{r}} \:{and}\:{D}_{{r}} \:{by}\:{highest}\:{power}\right. \\ $$$$\left.{n}\:\:{to}\:{get}\:\mathrm{0}\left({zero}\right)\:\:\:{because}\:\:\:\frac{\mathrm{1}}{\infty}\rightarrow\mathrm{0}\right] \\ $$
Commented by Rio Michael last updated on 24/Jul/19
$${thanks},\:{at}\:{times}\:{i}\:{just}\:{think}\:{of}\:{infinity}\:{as}\:{a}\:{number} \\ $$