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

Explanation:

Step1: For part a, divide numerator and denominator by $x^{2}$

When $x\to\infty$, we have $\lim_{x\to\infty}\frac{9x^{2}-4x + 2}{3x^{2}-1}=\lim_{x\to\infty}\frac{9-\frac{4}{x}+\frac{2}{x^{2}}}{3-\frac{1}{x^{2}}}$. As $x\to\infty$, $\frac{4}{x}\to0$, $\frac{2}{x^{2}}\to0$ and $\frac{1}{x^{2}}\to0$. So the limit is $\frac{9 - 0+0}{3-0}=3$.

Step2: For part b, divide numerator and denominator by $x^{2}$

When $x\to-\infty$, $\lim_{x\to-\infty}\frac{-4x + 1}{3x^{2}-2}=\lim_{x\to-\infty}\frac{-\frac{4}{x}+\frac{1}{x^{2}}}{3-\frac{2}{x^{2}}}$. As $x\to-\infty$, $\frac{4}{x}\to0$, $\frac{1}{x^{2}}\to0$ and $\frac{2}{x^{2}}\to0$. So the limit is $\frac{0 + 0}{3-0}=0$.

Step3: For part c, divide numerator and denominator by $x^{2}$

When $x\to\infty$, $\lim_{x\to\infty}\frac{5x^{3}-2x - 3}{x^{2}-4}=\lim_{x\to\infty}\frac{5x-\frac{2}{x}-\frac{3}{x^{2}}}{1-\frac{4}{x^{2}}}$. As $x\to\infty$, $\frac{2}{x}\to0$, $\frac{3}{x^{2}}\to0$ and $\frac{4}{x^{2}}\to0$. The limit is $\infty$ since the numerator goes to $\infty$ as $x\to\infty$.

Step4: For part d, divide numerator and denominator by $x$

When $x\to\infty$, $\lim_{x\to\infty}\frac{\sqrt{5x^{2}-3}}{x + 2}=\lim_{x\to\infty}\frac{\sqrt{5-\frac{3}{x^{2}}}}{1+\frac{2}{x}}$. As $x\to\infty$, $\frac{3}{x^{2}}\to0$ and $\frac{2}{x}\to0$. So the limit is $\sqrt{5}$.

Step5: For part e, divide numerator and denominator inside the square - root by $x^{2}$

When $x\to\infty$, $\lim_{x\to\infty}\sqrt{\frac{8x^{2}-7x + 1}{4x^{2}-3x + 2}}=\sqrt{\lim_{x\to\infty}\frac{8-\frac{7}{x}+\frac{1}{x^{2}}}{4-\frac{3}{x}+\frac{2}{x^{2}}}}$. As $x\to\infty$, $\frac{7}{x}\to0$, $\frac{1}{x^{2}}\to0$, $\frac{3}{x}\to0$ and $\frac{2}{x^{2}}\to0$. So the limit is $\sqrt{2}$.

Answer:

a. $3$ b. $0$ c. $\infty$ d. $\sqrt{5}$ e. $\sqrt{2}$