Practice Problems for Module 3
Exercises
Directions: You should try to solve each problem first, and then click "Reveal Answer" to check your answer. You can click "Watch Video" if you need help with a problem.
1. Evaluate the limit.
- \(\displaystyle \lim_{x\rightarrow -3}\frac{x^2-x-12}{x+3}\)
- \(\displaystyle \lim_{x\rightarrow 1}\frac{x^2-x-2}{x+1}\)
- \(\displaystyle \lim_{t\rightarrow 0}\frac{\sqrt{2-t}-\sqrt{2}}{t}\)
- \(\displaystyle \lim_{h\rightarrow 0}\frac{(3+h)^{-1}-3^{-1}}{h}\)
- \(\displaystyle \lim_{t\rightarrow 1}\left\langle 2t-3, \frac{t^2-t}{t-1}\right\rangle\)
- \(-7\)
- \(-1\)
- \(-\dfrac{1}{2\sqrt{2}}\)
- \(-\dfrac{1}{9}\)
- \(\langle -1,1\rangle\)
2. Find the limit.
- \(\displaystyle \lim_{x\rightarrow -4^-}\frac{|x+4|}{x+4}\)
- \(\displaystyle \lim_{x\rightarrow -4^+}\frac{|x+4|}{x+4}\)
- \(\displaystyle \lim_{x\rightarrow -4}\frac{|x+4|}{x+4}\)
- \(-1\)
- \(1\)
- DNE
3. For the function below, evaluate each of the following limits if it exists.
\[f(x)=\left\{
\begin{array}{ll}
x & \textrm{if } x<0\\
x^2 & \textrm{if } 0<x\leq 2\\
8-x & \textrm{if } x>2\\
\end{array}\right.
\]
- \(\displaystyle \lim_{x\rightarrow 0^+} f(x)\)
- \(\displaystyle \lim_{x\rightarrow 0^-} f(x)\)
- \(\displaystyle \lim_{x\rightarrow 0} f(x)\)
- \(\displaystyle \lim_{x\rightarrow 1} f(x)\)
- \(\displaystyle \lim_{x\rightarrow 2^-} f(x)\)
- \(\displaystyle \lim_{x\rightarrow 2^+} f(x)\)
- \(\displaystyle \lim_{x\rightarrow 2} f(x)\)
- \(0\)
- \(0\)
- \(0\)
- \(1\)
- \(4\)
- \(6\)
- DNE
4. Explain why the following function is discontinuous at \(x=0\).
\[f(x)=\left\{
\begin{array}{ll}
\cos x & \textrm{if } x<0\\
0 & \textrm{if } x=0\\
1-x^2 & \textrm{if } x>0\\
\end{array}\right.
\]
5. Find the values of \(a\) and \(b\) that make \(f\) continuous everywhere.
\[f(x)=\left\{
\begin{array}{ll}
\dfrac{x^2-4}{x-2} & \textrm{if } x<2\\
ax^2-bx+3 & \textrm{if } 2\leq x <3\\
2x-a+b & \textrm{if } x\geq 3\\
\end{array}\right.
\]
6. Find the limit.
- \(\displaystyle \lim_{x\rightarrow \infty} \dfrac{1-x^2}{x^3-x+1}\)
- \(\displaystyle \lim_{x\rightarrow \infty} \dfrac{4x^3+6x^2-2}{2x^3-4x+5}\)
- \(\displaystyle \lim_{x\rightarrow \infty} \left( \sqrt{9x^2+x}-3x\right)\)
- \(\displaystyle \lim_{x\rightarrow -\infty} \dfrac{1+x^6}{x^4+1}\)
- \(\displaystyle \lim_{x\rightarrow- \infty} \dfrac{\sqrt{1+4x^6}}{2-x^3}\)
- \(\displaystyle \lim_{x\rightarrow \infty} \dfrac{\sqrt{x+3x^2}}{4x-1}\)
- \(\displaystyle \lim_{x\rightarrow \infty} \dfrac{e^{3x}-e^{-3x}}{e^{3x}+e^{-3x}}\)
- \(\displaystyle \lim_{x\rightarrow -\infty} \dfrac{e^{3x}-e^{-3x}}{e^{3x}+e^{-3x}}\)
- 0
- 2
- \(\dfrac{1}{6}\)
- \(+\infty\)
- 2
- \(\dfrac{\sqrt{3}}{4}\)
- 1
- \(-1\)