Question

ma 241 calculus 1
practice for test 1
fall 2025

1. find the limit. you must show how to take the limit using techniques in class! no shortcuts will earn points!

a. $lim_{x
ightarrowinfty}\frac{9x^{2}-4x + 2}{3x^{2}-1}=$
c. $lim_{x
ightarrow+infty}\frac{5x^{3}-2x - 3}{x^{2}-4}=$
e. $lim_{x
ightarrowinfty}sqrt{\frac{8x^{2}-7x + 1}{4x^{2}-3x + 2}}=$
b. $lim_{x
ightarrow-infty}\frac{-4x + 1}{3x^{2}-2}=$
d. $lim_{x
ightarrowinfty}\frac{sqrt{5x^{2}-3}}{x + 2}=$

1. find the general derivative of $f(x)=\frac{1}{x + 2}$ using the limit definition. leave in terms of x. then find the tangent

Explanation:

a.

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

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

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:

$3$

b.