ma 241 calculus 1\npractice for test 1\nfall 2025\n26. find the limit. you must show how to take the limit…

a.

Explanation:

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

When $x\to\infty$, we have $\lim_{x\to\infty}\frac{9x^{2}-4x + 2}{3x^{2}-1}=\lim_{x\to\infty}\frac{9-\frac{4}{x}+\frac{2}{x^{2}}}{3-\frac{1}{x^{2}}}$

Step2: Apply limit rules

As $\lim_{x\to\infty}\frac{1}{x}=0$ and $\lim_{x\to\infty}\frac{1}{x^{2}} = 0$, we get $\frac{\lim_{x\to\infty}(9)-\lim_{x\to\infty}(\frac{4}{x})+\lim_{x\to\infty}(\frac{2}{x^{2}})}{\lim_{x\to\infty}(3)-\lim_{x\to\infty}(\frac{1}{x^{2}})}=\frac{9 - 0+0}{3-0}=3$

Answer:

$3$

b.

Explanation:

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

When $x\to-\infty$, $\lim_{x\to-\infty}\frac{-4x + 1}{3x^{2}-2}=\lim_{x\to-\infty}\frac{-\frac{4}{x}+\frac{1}{x^{2}}}{3-\frac{2}{x^{2}}}$

Step2: Apply limit rules

Since $\lim_{x\to-\infty}\frac{1}{x}=0$ and $\lim_{x\to-\infty}\frac{1}{x^{2}}=0$, we have $\frac{\lim_{x\to-\infty}(-\frac{4}{x})+\lim_{x\to-\infty}(\frac{1}{x^{2}})}{\lim_{x\to-\infty}(3)-\lim_{x\to-\infty}(\frac{2}{x^{2}})}=\frac{0 + 0}{3-0}=0$

Answer:

$0$

c.

Explanation:

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

For $\lim_{x\to+\infty}\frac{5x^{3}-2x - 3}{x^{2}-4}$, we rewrite it as $\lim_{x\to+\infty}\frac{5x-\frac{2}{x}-\frac{3}{x^{2}}}{1-\frac{4}{x^{2}}}$

Step2: Apply limit rules

As $\lim_{x\to+\infty}\frac{1}{x}=0$ and $\lim_{x\to+\infty}\frac{1}{x^{2}}=0$, we get $\lim_{x\to+\infty}(5x-\frac{2}{x}-\frac{3}{x^{2}})\to+\infty$ and $\lim_{x\to+\infty}(1 - \frac{4}{x^{2}})=1$, so the limit is $+\infty$

Answer:

$+\infty$

d.

Explanation:

Step1: Divide numerator and denominator by $x$

For $\lim_{x\to+\infty}\frac{\sqrt{5x^{2}-3}}{x + 2}$, rewrite it as $\lim_{x\to+\infty}\frac{\sqrt{5-\frac{3}{x^{2}}}}{1+\frac{2}{x}}$ (since for $x>0$, $\frac{\sqrt{5x^{2}-3}}{x}=\sqrt{\frac{5x^{2}-3}{x^{2}}}=\sqrt{5-\frac{3}{x^{2}}}$)

Step2: Apply limit rules

As $\lim_{x\to+\infty}\frac{1}{x}=0$ and $\lim_{x\to+\infty}\frac{1}{x^{2}}=0$, we have $\frac{\lim_{x\to+\infty}\sqrt{5-\frac{3}{x^{2}}}}{\lim_{x\to+\infty}(1+\frac{2}{x})}=\frac{\sqrt{5-0}}{1 + 0}=\sqrt{5}$

Answer:

$\sqrt{5}$

e.

Explanation:

Step1: Divide numerator and denominator inside the square - root by $x^{2}$

$\lim_{x\to+\infty}\sqrt{\frac{8x^{2}-7x + 1}{4x^{2}-3x + 2}}=\lim_{x\to+\infty}\sqrt{\frac{8-\frac{7}{x}+\frac{1}{x^{2}}}{4-\frac{3}{x}+\frac{2}{x^{2}}}}$

Step2: Apply limit rules

Since $\lim_{x\to+\infty}\frac{1}{x}=0$ and $\lim_{x\to+\infty}\frac{1}{x^{2}}=0$, we get $\sqrt{\frac{\lim_{x\to+\infty}(8)-\lim_{x\to+\infty}(\frac{7}{x})+\lim_{x\to+\infty}(\frac{1}{x^{2}})}{\lim_{x\to+\infty}(4)-\lim_{x\to+\infty}(\frac{3}{x})+\lim_{x\to+\infty}(\frac{2}{x^{2}})}}=\sqrt{\frac{8-0 + 0}{4-0+0}}=\sqrt{2}$

Answer:

$\sqrt{2}$