As we move on to Chapter 3, the theme of our work will focus on how we can apply the meaning of the derivative to solve important problems. Interestingly, although we use limits to define the derivative itself, it turns out that the derivative can be a useful tool in evaluating challenging limits of a certain type.
The topics discussed in this section are: Indeterminate forms of type β0/0β. Local linearization and LβHΓ΄pitalβs Rule. Infinite limits and limits at infinity. Asymptotes. Indeterminate forms of type β\(\infty/\infty\)β. Indeterminate products of the form β\(0 \cdot \infty\)β. Indeterminate powers of the form β\(\infty^{0}\)β, β\(1^{\infty}\)β, β\(0^{0}\)β.
These are the tasks you should be able to perform with reasonable fluency when you arrive at our next class meeting. Important new vocabulary words are indicated in italics.
π [Submit] Do the following construction (see FigureΒ 75) in GeoGebra. Note how the red graph of the quotient \(h(x)\) suggests that \(\displaystyle \lim_{x\rightarrow 1}\frac{x^{5}+x-2}{x^{2}-1}= 3\text{.}\) The value of \(b\) is what is computed via LβHΓ΄pitalβs rule.
Repeat this construction to analyze the limit found in Activity 3.2.2(a). That is, use \(f(x) = \ln(1+x)\text{,}\)\(g(x) = x\text{,}\) and investigate the limit of \(f(x)/g(x)\) as \(x \rightarrow 0\text{.}\) Submit screenshots as needed.
π [Submit] Do form a table of values using a spreadsheet (Excel or Google Sheets) to form a hypothesis regarding the value of \(\displaystyle \lim_{x \rightarrow 0}\frac{e^{2x}-1}{x}\text{.}\) Values of \(x\) that are both negative and positive near zero should be used. Then, repeat the GeoGebra exploration above to see if the same value for the limit emerges. [Yes, you should get 2.] Sample output is shown in FigureΒ 76. Submit screenshots as needed.
We attempt to evaluate the value of \(\displaystyle \lim_{x \rightarrow 0}\frac{e^{2x}-1}{x}\text{.}\) Define \(f(x) = e^{2x}-1\) (the numerator) and define \(g(x) = x\) (the denominator). Then \(\displaystyle \lim_{x \rightarrow 0}\frac{e^{2x}-1}{x}= \lim_{x \rightarrow 0}\frac{f(x)}{g(x)}.\)
As \(x \rightarrow 0\text{,}\) the approximation becomes better. This tells us \(\displaystyle \lim_{x \rightarrow 0}\frac{e^{2x}-1}{x}={\underline{\hspace{30mm}}}\text{.}\)
Find limits involving indeterminate powers (β\(\infty^{0}\)β, β\(1^{\infty}\)β, β\(0^{0}\)β): use the identity \(u=e^{\ln u}\) to rewrite the expression and find the limit of the exponent (you may get an indeterminate power and LβHΓ΄pitalβs Rule might be required at this point).
Realize which forms of limits are not indeterminate, i.e. which always result in a clear answer (e.g. β\(0^{\infty}\)β=0, β\(*/\infty\)β=0, β\(*/0^{+}\)β=\(\infty\))
Use the graphs of \(f\) and \(g\) and their tangent lines at \((2,0)\) shown in FigureΒ 79 to find \(\displaystyle \lim_{x \rightarrow 2}\frac{f(x)}{g(x)}\text{.}\)
Calculate \(\displaystyle \lim_{x \rightarrow \infty}\left( 1 + \frac{1}{x}\right)^{x}\text{.}\)Hint: Note that this limit has the form \(1^{\infty}\text{.}\) So write
Calculate \(\displaystyle \lim_{x \rightarrow \infty}xe^{-x}\text{.}\)Hint: Note that this limit has the form \(\infty \cdot 0\text{.}\) Rewrite \(\displaystyle xe^{-x}= \frac{x}{e^{x}}\) so that the limit takes the form \(\displaystyle \frac{\infty}{\infty}\text{.}\)