A Modeling Approach to Calculus: MA2C

1. The sequence \(x_n = 3-4n\) is recognized as the explicit formula of an arithmetic sequence with increment \(c=-4\text{.}\) Since this is a negative increment, the sequence decreases without bound. So we write

   \begin{equation*} \lim_{n \to \infty} 3-4n = -\infty. \end{equation*}

2. The sequence \(x_n = 100+0.02n\) is arithmetic with increment \(c=0.02\text{.}\) Since the increment is positive, the sequence increases without bound and we write

   \begin{equation*} \lim_{n \to \infty} 100+0.02n = +\infty. \end{equation*}

3. The sequence \(x_n=-3 \cdot 1.05^n\) is a geometric sequence with ratio \(\rho=1.05\text{.}\) Because \(\rho \gt 1\text{,}\) the sequence grows without bound. Furthermore, because the terms are all negative, we have a limit

   \begin{equation*} \lim_{n \to \infty} -3 \cdot 1.05^n = -\infty. \end{equation*}

4. The sequence \(x_n = 100 \cdot (-0.75)^n\) is a geometric sequence with ratio \(\rho=-0.75\text{.}\) Because the ratio is negative, the signs of the terms alternate between positive and negative. However, since \(|\rho|= 0.75 \lt 1\text{,}\) the magnitude of the terms converges to zero so that

   \begin{equation*} \lim_{n \to \infty} 100 \cdot (-0.75)^n = 0. \end{equation*}

5. The sequence \(x_n = 5 \cdot (-1.5)^n\) has a negative ratio \(\rho=-1.5\text{.}\) Since \(|\rho| \gt 1\text{,}\) the terms have alternating signs but grow in magnitude. Consequently, \(\displaystyle \lim_{n \to \infty} 5 \cdot (-1.5)^n\) does not exist.

1. The sequence \(x_n = 3 + \frac{5}{2^n}\) is the sum of terms \(3\) and \(\frac{5}{2^n}\text{.}\) The constant sequence has a limit \(3 \to 3\) (since it never changes). The geometric sequence \(\frac{5}{2^n} = 5 \cdot (\frac{1}{2})^n\) has a ratio \(|\rho| \lt 1\) so that \(\frac{5}{2^n} \to 0\text{.}\) The form of the limit (using the terms) is

   \begin{equation*} \lim_{n \to \infty} 3 + \frac{5}{2^n} = 3 + 0, \end{equation*}

   and since this does not involve any cancelation of infinities, will give the correct limit,

   \begin{equation*} \lim_{n \to \infty} 3+\frac{5}{2^n} = 3+0 = 3. \end{equation*}

2. The sequence \(\displaystyle u_n = \frac{n^2-3n}{5n-1}\) is a quotient of terms \(n^2-3n\) and \(5n-1\text{.}\) To find the limit, we explore the terms individually first.

   Because \(n^2 = n \cdot n\text{,}\) we know \(n^2 \to +\infty \cdot +\infty = +\infty\text{.}\) Similarly, the arithmetic sequence \(3n \to +\infty\text{.}\) However, the difference \(n^2 - 3n\) would have a limit of the form \(+\infty - \infty\text{,}\) which is a cancellation of infinities. As written, \(n^2-3n\) is an indeterminate form.

   Our strategy will be to rewrite this as a product, and the best practice is to factor out (divide out) the greatest power of \(n\) (dominant term),

   \begin{equation*} n^2-3n = n^2(\frac{n^2}{n^2} - \frac{3n}{n^2}) = n^2 (1 - \frac{3}{n}). \end{equation*}

   From this, we find

   \begin{equation*} \frac{3}{n} \to \frac{3}{+\infty} = 0 \quad \Rightarrow \quad 1-\frac{3}{n} \to 1-0 = 1. \end{equation*}

   Since we already know \(n^2 \to +\infty\text{,}\) we have the limit of the numerator

   \begin{equation*} \lim_{n \to \infty} n^2-3n = \lim_{n \to \infty} n^2(1-\frac{3}{n}) = +\infty \cdot 1 = +\infty. \end{equation*}

   The term in the denominator \(5n-1\) is an arithmetic sequence (linear function) with increment (slope) \(c=5\text{.}\) We know

   \begin{equation*} \lim_{n \to \infty} 5n-1 = +\infty. \end{equation*}

   Unfortunately, that means our limit form as a quotient is itself an indeterminate form,

   \begin{equation*} \lim_{n \to \infty} \frac{n^2-3n}{5n-1} = \frac{+\infty}{+\infty}. \end{equation*}

   We can not cancel infinities, so we must rewrite our formula.

   For this example, I worked out the numerator separately to make a point about that term itself being an indeterminate form. In practice, our strategy will be to apply that factoring principle to the entire formula at one step. This is illustrated below.

   The problem can be solved up if we just factor out from the numerator and denominator the dominant term (greatest power) and simplify as needed.

   \begin{align*} \lim_{n \to \infty} \frac{n^2-3n}{5n-1} & = \lim_{n \to \infty} \frac{n^2(1-\frac{3}{n})}{n(5-\frac{1}{n})} \\ & = \lim_{n \to \infty} \frac{n(1-\frac{3}{n})}{5-\frac{1}{n}} \\ & = \frac{+\infty \cdot(1-\frac{3}{\infty})}{5-\frac{1}{\infty}}\\ & = \frac{+\infty\cdot 1}{5} = +\infty. \end{align*}

3. The sequence \(\displaystyle w_n = \frac{1+2^n}{3+5^n}\text{,}\) by quick inspection, involves the geometric sequences \(2^n\) and \(5^n\text{,}\) both of which grow exponentially so that \(w_n \to \frac{+\infty}{+\infty}\text{.}\) This indeterminate form involves canceling infinities, so we must rewrite the formula. Following the method of the previous example, we factor out the dominant term, in this case the geometrically growing powers.

   \begin{align*} \lim_{n \to \infty} \frac{1+2^n}{3+5^n} & = \lim_{n \to \infty} \frac{2^n(\frac{1}{2^n} + \frac{2^n}{2^n})}{5^n(\frac{3}{5^n}+\frac{5^n}{5^n})} \\ & = \lim_{n \to \infty} \frac{2^n(\frac{1}{2^n}+1)}{5^n(\frac{3}{5^n}+1)} \end{align*}

   This is still indeterminate form \(\frac{+\infty}{+\infty}\text{,}\) so we rewrite \(\displaystyle \frac{2^n}{5^n} = ({\textstyle \frac{2}{5}})^n\text{,}\) which is a geometric sequence with ratio \(\rho=\frac{2}{5}\) satisfying \(|\rho| \lt 1\text{.}\)

   \begin{align*} \lim_{n \to \infty} \frac{1+2^n}{3+5^n} & = \lim_{n \to \infty} \frac{2^n(\frac{1}{2^n}+1)}{5^n(\frac{3}{5^n}+1)}\\ & = \lim_{n \to \infty} \frac{(\frac{2}{5})^n(\frac{1}{2^n}+1)}{\frac{3}{5^n}+1}\\ & = \frac{0 \cdot (0+1)}{0+1} = \frac{0}{1} = 0 \end{align*}

A sequence that has a linear projection function, involving both multiplication by a ratio and the addition of an increment, has a shifted geometric sequence as its explicit formula. The equilibrium value is found by solving the fixed point equation.

Thus, the equilibrium value is \(u^* = 100\text{.}\)

The explicit formula is a geometric sequence with ratio \(\alpha=0.8\) shifted by the equilibrium,

Be careful not to violate the order of operations by adding the \(100\) and \(-50\) or multiplying the \(-50\) and \(0.8\text{,}\)

Using this explicit formula, we can find the limit of the sequence. The geometric sequence has a limit 0 because the ratio has magnitude smaller than 1.

That is, the sequence converges to the equilibrium value.

Find the fixed point by solving the equation \(x=1.05x - 20\text{.}\)

Using the fixed point \(x^*=400\) and the growth factor \(\alpha=1.05\text{,}\) we can write down the explicit formula,

The geometric term with ratio \(\alpha=1.05\) grows without bound. The limit of the sequence can be found:

The sequence will decrease without bound.

The patient's body starts with no drug. Immediately after the first dose, there are 500 mg. Four hours later, 60% has been removed and then another dose is added in. If we let \(D_n\) be the sequence of drug mass in the body as a function of the number of doses \(n\text{,}\) then this is modeled recursively by the equation

with initial value \(D_1 = 500\text{.}\)

This model has a linear projection function \(f(x) = 0.4x+500\) and corresponding fixed point

The explicit formula, using Theorem 4.3.7, is given by

Because the slope of the projection function \(\alpha = 0.4\) has magnitude less than 1, the limiting value is the fixed point \(x^* = \frac{2500}{3} \approx 833.33\text{.}\) Thus, if the patient continues to take the drug, the amount in the body immediately after each dose will be approximately 833.33 mg. (Immediately before the dose, it must have been approximately 333.33 mg.)

There are two major steps needed for this problem: (i) find an explicit formula for the summation that depends only on \(n\) and (ii) compute the limit of that explicit formula.

To compute the explicit formula for the summation, we need to remember that the variable \(n\) is a constant with respect to the summation index \(k\text{.}\) We will rewrite the formula of the sequence in summation to be a sum so that we can use the linearity property.

We see that the summation is itself a sequence involving an index \(n\text{.}\) We find the limit of that sequence. The fraction \(\displaystyle \frac{9n+9}{2n}\) will be indeterminate form \(\frac{\infty}{\infty}\text{,}\) so we will factor out the dominant term of \(n\) from top and bottom.