Question

evaluate the limit $lim_{x
ightarrowinfty}\frac{9x^{2}-8x + 3}{5x+4}$

Explanation:

Step1: Divide numerator and denominator by $x^2$

When $x\to\infty$, we have $\lim_{x\to\infty}\frac{9x^{2}-8x + 3}{5x+4}=\lim_{x\to\infty}\frac{9-\frac{8}{x}+\frac{3}{x^{2}}}{\frac{5}{x}+\frac{4}{x^{2}}}$

Step2: Use limit properties

We know that $\lim_{x\to\infty}\frac{1}{x}=0$ and $\lim_{x\to\infty}\frac{1}{x^{2}} = 0$. So, $\lim_{x\to\infty}\frac{9-\frac{8}{x}+\frac{3}{x^{2}}}{\frac{5}{x}+\frac{4}{x^{2}}}=\frac{\lim_{x\to\infty}(9)-\lim_{x\to\infty}(\frac{8}{x})+\lim_{x\to\infty}(\frac{3}{x^{2}})}{\lim_{x\to\infty}(\frac{5}{x})+\lim_{x\to\infty}(\frac{4}{x^{2}})}$
Since $\lim_{x\to\infty}(9) = 9$, $\lim_{x\to\infty}\frac{8}{x}=0$, $\lim_{x\to\infty}\frac{3}{x^{2}}=0$, $\lim_{x\to\infty}\frac{5}{x}=0$ and $\lim_{x\to\infty}\frac{4}{x^{2}}=0$, we get $\frac{9 - 0+0}{0 + 0}=\infty$

Answer:

$\infty$