Question

let $h(x)=\frac{x^{2}-3}{x^{4}+9}$. find $lim_{x
ightarrowinfty}h(x)$. choose 1 answer: a 1 b 0 c $-\frac{1}{3}$ d the limit is unbounded

Explanation:

Step1: Divide numerator and denominator by $x^4$

$\lim_{x
ightarrow\infty}\frac{x^{2}-3}{x^{4}+9}=\lim_{x
ightarrow\infty}\frac{\frac{x^{2}}{x^{4}}-\frac{3}{x^{4}}}{\frac{x^{4}}{x^{4}}+\frac{9}{x^{4}}}=\lim_{x
ightarrow\infty}\frac{\frac{1}{x^{2}}-\frac{3}{x^{4}}}{1 + \frac{9}{x^{4}}}$

Step2: Use limit properties

We know that $\lim_{x
ightarrow\infty}\frac{1}{x^{n}} = 0$ for $n>0$. So, $\lim_{x
ightarrow\infty}\frac{1}{x^{2}}=0$, $\lim_{x
ightarrow\infty}\frac{3}{x^{4}} = 0$ and $\lim_{x
ightarrow\infty}\frac{9}{x^{4}}=0$. Then $\lim_{x
ightarrow\infty}\frac{\frac{1}{x^{2}}-\frac{3}{x^{4}}}{1+\frac{9}{x^{4}}}=\frac{0 - 0}{1+0}=0$

Answer:

B. 0