Lab 04 - Limits, Infinity, and Asymptotes.ipynb

Kernel: SageMath 9.2

Lab 04 - Limits, Inifinity, and Asymptotes

Overview

The concept of a limit is the central idea in Calculus. Limits involving infininity are important because they are closely related to asymptotes. While asymptotes for functions are sometimes easy to identify from a graph, the actual definition of asymptotes is in terms of limits. There are several types of asymptotes and the two simplest are:

ParseError: KaTeX parse error: Undefined control sequence: \hfill at position 31: …l|l|l|} \hline \̲h̲f̲i̲l̲l̲ ̲\textbf{Asympto…

In this lab, we will use SageMath to evaluate the limit of a function at a point and we will use limits to identify asymptotes.

Important SageMath Commands Introduced in this Lab

ParseError: KaTeX parse error: Undefined control sequence: \hfill at position 32: …|l|l|} \hline \̲h̲f̲i̲l̲l̲ ̲\textbf{Command…

Related Course Material

Sections 2.1, 2.2, 2.4, and 2.6

Example 1

Use SageMath to evaluate $\displaystyle \lim_{h\rightarrow 0} \dfrac{f(x+h) - f(x)}{h}$ for the functions below. Don't forget that you must make $h$ a variable before you can use it as one.

1. $f(x) = x^2$

2. $f(x) = \sqrt{x}$

3. $f(x) = \frac{1}{x}$

4. $f(x) = \sin(x)$

5. $f(x) = e^x$ (Depending on your version of SageMath, you may need to add the line $\textbf{assume(x, 'noninteger')}$ in order for SageMath to compute this limit)

Example 2

Let $f(x) = \frac{|x|}{x}$ and use SageMath to calculate the following limits. You can use abs($x$) command for the absolute value function.

1. $\displaystyle \lim_{x\rightarrow 5} f(x)$

2. $\displaystyle \lim_{x\rightarrow -\infty} f(x)$

3. $\displaystyle \lim_{x\rightarrow 0} f(x)$

4. $\displaystyle \lim_{x\rightarrow 0^+} f(x)$

5. $\displaystyle \lim_{x\rightarrow 0^-} f(x)$

Based off of your results, what do you think the graph of $f(x)$ looks like? Check your guess below.

Example 3

Now, we will use SageMath to identify the asymptotes of a function. Let $f(x) = \dfrac{5 + 2x^2}{x^2 - 1}$. Recall that the values which make the denominator of a rational function equal zero are all of the possible places for a vertical asymptote. We can find all $x$-values which make the denominator of $f(x)$ equal zero by hand or we can use SageMath to help us. If we use SageMath, we need to use the functions $\textit{expression}\textbf{.denominator(normalize=False)}$ and $\textbf{solve}(\textit{equation}, \textit{variable})$. When creating an equation in SageMath, you must use $\textbf{==}$ in place of $\textbf{=}$.

def f(x):
    return (5 + 2*x^2)/(x^2-1)
solve(f(x).denominator(normalize=False) == 0,x)

SageMath either returned $[1, -1]$ or $[ ]$. We know that the answers to the equation should be $x = 1, -1$, so why did SageMath return $[ ]$ which means that there isn't an answer? This is because earlier in Example 1, part 5 we added the condition that $x$ is a noninteger in order to compute the limit. Therefore, SageMath still thinks that SageMath is a noninteger and is not allowed to be $1$ or $-1$. We can remove the assumption on $x$ by using the $\textbf{forget()}$ command.

forget()
solve(f(x).denominator(normalize=False) == 0,x)

SageMath tells us that the only $x$-values which make the denomiator equal zero are $x = -1$ and $x = 1$. Therefore, the only possible vertical asymptotes are at $x=-1$ and $x=1$. We can use SageMath to calculate the one-sided limits of $f(x)$ at these $x$-values to determine if either of them are actually vertical asymptotes.

limit(f(x), x=-1, dir='+')

limit(f(x), x=-1, dir='-')

Since at least one of the one-sided llimits at $x = -1$ was either $\infty$ or $-\infty$, we have that the graph of $f(x)$ has a vertical asymptote at $x = -1$.

Check the one sided limits at $x = 1$ to see if there is also a vertical asymptote at $x=1$.

Therefore, the graph of $f(x)$ has vertical asymptotes at both $x=-1$ and $x=1$.

To find the horizontal asymptotes, we calculate the limit of $f(x)$ as $x$ approaches $\infty$ and $-\infty$.

limit(f(x), x=infinity)

limit(f(x), x=-infinity)

Both limits tell us that $y = 2$ is a horizontal asymptote. Therefore, we have that $y = 2$ is our only horizontal asymptote.

We can check our conclusion by plotting the function in an appropriate viewing window which will show all of the asymptotes. The option ParseError: KaTeX parse error: Expected 'EOF', got '_' at position 15: \textbf{detect_̲poles = 'show'} will plot the vertical asymptote(s) on the graph. We can also plot the horizontal asymptote(s) by inclding the $y$-value(s) in our plot.

$\textbf{Note:}$ Sometimes, ParseError: KaTeX parse error: Expected 'EOF', got '_' at position 15: \textbf{detect_̲poles = 'show'} does not plot the vertical asymptote.

plot([f(x), 2], xmin=-10, xmax=10, ymin=-10, ymax=10, color = ['blue', 'gray'], 
     linestyle = ['-', '--'], detect_poles = 'show')

Example 4

Repeat Example 3 with the following functions:

1. $f(x) = \dfrac{3x + 2}{(x+2)^2}$

2. $f(x) = \dfrac{\sqrt{9x^2 + 4} - 2}{x + 1}$

3. $f(x) = \dfrac{(x^2 - 1)\sqrt{4x^2 + 1}}{x^3 - 2x^2 - x + 2}$

4. $f(x) = \dfrac{\sin(x)}{x}$