Question

the limit found using either method is $limlimits_{x\to+infty}\frac{5x^{2}-7x}{9x^{2}+4}=square$. (simplify your answer.)

Explanation:

Step1: Divide numerator and denominator by \(x^{2}\)

$$\lim_{x ightarrow\infty}\frac{5x^{2}-7x}{9x^{2}+4}=\lim_{x ightarrow\infty}\frac{\frac{5x^{2}}{x^{2}}-\frac{7x}{x^{2}}}{\frac{9x^{2}}{x^{2}}+\frac{4}{x^{2}}}$$

Step2: Simplify the expression

$$=\lim_{x ightarrow\infty}\frac{5 - \frac{7}{x}}{9+\frac{4}{x^{2}}}$$

Step3: Use the limit property \(\lim_{x

ightarrow\infty}\frac{1}{x^{n}} = 0\) (\(n>0\))
As \(x
ightarrow\infty\), \(\lim_{x
ightarrow\infty}\frac{7}{x}=0\) and \(\lim_{x
ightarrow\infty}\frac{4}{x^{2}} = 0\). So the limit becomes \(\frac{5-0}{9 + 0}\)

Answer:

\(\frac{5}{9}\)