Question

Evaluate the limit of

Answer 1

Answer:

Option a is correct -- $\frac{9}{2}$

Step-by-step explanation:

If we put infinity directly into the expression we get ∞ + ∞ expression. In order to circumvent it we divide both numerator and denominator with the greatest exponent.

$\lim_{x \to \infty} \frac{9n^{3}+5n-2}{2n^{3}}$

 Divide each term by n³

$=\lim_{x \to \infty} \frac{\frac{9n^{3}}{n^{3}}+\frac{5n}{n^{3}} -\frac{2}{n^{3}}}{\frac{2n^{3}}{n^{3}} }$

$=\lim_{x \to \infty}\frac{9(1)+\frac{5}{n^{2}}-\frac{2}{n^{3}}}{2(1)} } \\=\lim_{x \to \infty}\frac{9+\frac{5}{n^{2}}-\frac{2}{n^{3}}}{2}} \\\\by putting n = infinity \frac{5}{n^{2}} becomes 0 and\frac{2}{n^{3}} becomes 0 \\=\frac{9+0-0}{2} \\=\frac{9}{2}$