Tìm a để lim (a.n^2+4n)/(8n^2+3) = 3/4

* Hướng dẫn giải

$\lim \frac{\mathrm{a}.{\mathrm{n}}^2+4\mathrm{n}}{8{\mathrm{n}}^2+3}=\lim \frac{\mathrm{a}+\frac{4}{\mathrm{n}}}{8+\frac{3}{{\mathrm{n}}^2}}$$=\frac{\lim \left(\mathrm{a}+\frac{4}{\mathrm{n}}\right)}{\lim \left(8+\frac{3}{{\mathrm{n}}^2}\right)}$$=\frac{\mathrm{a}}{8}.$

$\lim \frac{\mathrm{a}.{\mathrm{n}}^2+4\mathrm{n}}{8{\mathrm{n}}^2+3}=\frac{3}{4}\iff \frac{\mathrm{a}}{8}=\frac{3}{4}$$\iff \mathrm{a}=6.$