Question

your turn now:
x gets larger, we get closer to what value?
$3+\frac{1}{x}approxapprox$
$-2-\frac{32}{x}approxapprox$
$\frac{1}{2}+\frac{5}{x^{2}}approxapprox$

Explanation:

Step1: Analyze the limit of \(\frac{1}{x}\) as \(x\) gets larger

As \(x\to+\infty\), \(\lim_{x\to+\infty}\frac{1}{x} = 0\). For the expression \(3+\frac{1}{x}\), when \(x\) is very large, \(\frac{1}{x}\approx0\), so \(3+\frac{1}{x}\approx3 + 0=3\).

Step2: Analyze the limit of \(\frac{32}{x}\) as \(x\) gets larger

As \(x\to+\infty\), \(\lim_{x\to+\infty}\frac{32}{x}=0\). For the expression \(- 2-\frac{32}{x}\), when \(x\) is very large, \(\frac{32}{x}\approx0\), so \(-2-\frac{32}{x}\approx-2-0=-2\).

Step3: Analyze the limit of \(\frac{5}{x^{2}}\) as \(x\) gets larger

As \(x\to+\infty\), \(\lim_{x\to+\infty}\frac{5}{x^{2}} = 0\). For the expression \(\frac{1}{2}+\frac{5}{x^{2}}\), when \(x\) is very large, \(\frac{5}{x^{2}}\approx0\), so \(\frac{1}{2}+\frac{5}{x^{2}}\approx\frac{1}{2}+0=\frac{1}{2}\).

Answer:

For \(3+\frac{1}{x}\), the value is \(3\); for \(-2-\frac{32}{x}\), the value is \(-2\); for \(\frac{1}{2}+\frac{5}{x^{2}}\), the value is \(\frac{1}{2}\)