Problem 1 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

a. Find a number, \(\delta>0,\) so that if \(x\) is a number and \(0<|x-2|<\delta\) then \(|2 x-4|<0.01\). b. Find a number, \(\delta\), so that if \(x\) is a number and \(0<|x-2|<\delta\) then \(\left|x^{2}-4\right|<0.01\). c. Find a number, \(\delta\), so that if \(x\) is a number and \(0<|x-2|<\delta\) then \(\left|\frac{1}{x}-\frac{1}{2}\right|<0.01\). d. Find a number, \(\delta,\) so that if \(x\) is a number and \(0<|x-2|<\delta\) then \(\left|x^{3}-8\right|<0.01\). e. Find a number, \(\delta\), so that if \(x\) is a number and \(0<|x-2|<\delta\) then \(\left|\frac{x}{x+1}-\frac{2}{3}\right|<0.01\). f. Find a number, \(\delta>0\), so that if \(x\) is a number and \(0<|x-9|<\delta\) then \(|\sqrt{x}-3|<0.01\). g. Find a number, \(\delta\), so that if \(x\) is a number and \(0<|x-8|<\delta\) then \(|\sqrt[3]{x}-2|<0.01\). h. Find a number, \(\delta>0\), so that if \(x\) is a number and \(0<|x-2|<\delta\) then \(\left|x^{4}-3 x-13\right|<0.01\)

Step-by-Step Solution

Answer

(a) δ = 0.005, (b) δ = 0.0025, (c) δ = 0.04, (d) δ = 0.0008333, (e) δ = 0.09, (f) δ = 0.0601, (g) δ = 0.120601, (h) δ ≈ 0.001.

Step 1: Understanding Problem (a)

We need to ensure that if \( |x-2|<\delta \), then \( |2x-4| < 0.01 \). By simplifying \( |2x-4| \), this becomes \( 2|x-2| \). Thus, \( 2|x-2| < 0.01 \).

Step 2: Finding δ for Part (a)

We have \( 2|x-2| < 0.01 \), which implies \( |x-2| < 0.005 \). So, we can choose \( \delta = 0.005 \).

Step 3: Understanding Problem (b)

We need \( |x^{2}-4| < 0.01 \). Recognize that \( x^2 - 4 = (x-2)(x+2) \). Thus, \( |(x-2)(x+2)| < 0.01 \).

Step 4: Finding δ for Part (b)

Assume \( |x-2| < \delta \) and note that near \( x = 2 \), \( |x+2| \approx 4 \). Since \( |(x-2)(x+2)| < 0.01 \), \( |x-2| * 4 < 0.01 \) implies \( |x-2| < 0.0025 \). Therefore, \( \delta = 0.0025 \).

Step 5: Understanding Problem (c)

Simplify \( \left|\frac{1}{x} - \frac{1}{2}\right| < 0.01 \) to the condition \( \frac{|2-x|}{2x} < 0.01 \).

Step 6: Finding δ for Part (c)

For small \( |x-2| < \delta\), estimate \( x \) around 2. We get \( \left|\frac{x-2}{2x}\right| < 0.01 \). Using \( x \approx 2 \), then \( \left|\frac{x-2}{4}\right| < 0.01 \), so \( |x-2| < 0.04 \). Hence, \( \delta = 0.04 \).

Step 7: Understanding Problem (d)

Use \( |x^{3}-8| < 0.01 \), noting \( x^3 = 8 \) when \( x = 2 \). Close to \( x = 2 \), \( x^3 - 8 = (x-2)(x^2+2x+4) \) is significant.

Step 8: Finding δ for Part (d)

With \( |(x-2)(x^2+2x+4)| < 0.01 \), assuming \( x \approx 2 \), we have \( x^2 + 2x + 4 \approx 12 \). Then \( |x-2| < \frac{0.01}{12} \approx 0.0008333 \), so \( \delta = 0.0008333 \).

Step 9: Understanding Problem (e)

Start with \( \left|\frac{x}{x+1} - \frac{2}{3}\right| < 0.01 \). This simplifies to \( \left|\frac{3x-2(x+1)}{3(x+1)}\right| < 0.01 \), hence \( \left|\frac{x-2}{3(x+1)}\right| < 0.01 \).

Step 10: Finding δ for Part (e)

Assume \( x \approx 2 \), \( x+1 \approx 3 \), so \( \left|\frac{x-2}{9}\right| < 0.01 \). Then \( |x-2| < 0.09 \). Thus, \( \delta = 0.09 \).

Step 11: Understanding Problem (f)

We need \( |\sqrt{x} - 3| < 0.01 \), and note that if \( \sqrt{x} = 3 \), \( x = 9 \). So, solve \( |\sqrt{x}-3| < 0.01 \).

Step 12: Finding δ for Part (f)

Convert \( |\sqrt{x} - 3| < 0.01 \) to \( 3 - 0.01 < \sqrt{x} < 3 + 0.01 \), squared gives \( 8.9401 < x < 9.0601 \). Hence, \( |x-9| < 0.0601 \). Thus, \( \delta = 0.0601 \).

Step 13: Understanding Problem (g)

We need \( |\sqrt[3]{x} - 2| < 0.01 \) where \( \sqrt[3]{8} = 2 \). Solve \( 1.99 < \sqrt[3]{x} < 2.01 \).

Step 14: Finding δ for Part (g)

Cube \( 1.99 \) and \( 2.01 \) to get bounds: \( x \approx 7.880399 \) to \( x \approx 8.120601 \), so \( |x - 8| < 0.120601 \). Thus, \( \delta = 0.120601 \).

Step 15: Understanding Problem (h)

We need \( |x^{4} - 3x - 13| < 0.01 \). It's not easily factored like previous problems, so numeric exploration is needed for \( x \approx 2 \).

Step 16: Finding δ for Part (h)

Estimate \( x^{4} - 3x - 13 \) near \( x = 2 \). At \( x = 2 \), \( 2.01^{4} - 3 \cdot 2.01 - 13 < 0.01 \), giving bounds: try numerical exploration near 2, resulting in \( \delta \approx 0.001 \).

Key Concepts

Delta-epsilon definition

Delta-epsilon definition is a formal way to describe the concept of limits in calculus. This definition allows us to prove rigorously that a function approaches a specific value as the input gets close to a particular point. In simpler terms:

• The limit of a function as it approaches a point can be defined using precise values, called delta (\(\delta\)) and epsilon (\(\epsilon\)).
• For each positive \(\epsilon\), we find a corresponding \(\delta > 0\), so that whenever the input is within \(\delta\) of the point, the function's output is within \(\epsilon\) of the limit.

By using this method, we ensure that the function doesn’t just seem to reach close to a value, but does so within the tight bounds we set. It is like setting a tolerance for both the inputs and outputs to be within small error margins.

Inequalities in calculus

Inequalities in calculus often appear when solving limits or analyzing function behaviors. Working with inequalities helps us:

• Form statements that show the bounds of function behavior around a point.
• Simplify expressions to find limits by bounding them from above and below.

For example, if we need to ensure that a condition like \(|x^2 - 4| < 0.01\) holds, we use inequalities:

• Start by expressing the function in a simpler form, such as \((x-2)(x+2)\).
• Estimate values close to the point of interest (like \(x = 2\) in the example), to simplify inequality manipulations and find suitable \(\delta\).

These steps help in practically approaching the concept of limits and calculating specific values of \(\delta\) effectively.

Continuous functions

Continuous functions are those that have no holes, jumps, or breaks in their graphs. This smoothness means:

• For any small change in the input (within some radius), there is a small change in the output.
• If a function is continuous at a point, its limit at that point equals its value there.

The delta-epsilon definition can also be applied to understand continuity. When finding \(\delta\) from \(\epsilon\), it's assuming continuity around that point, suggesting outputs don’t change unexpectedly. For instance:

• Assume \(|\sqrt{x}-3| < 0.01\); here, continuity lets \(\sqrt{x}\) approach a well-defined \(\epsilon\). Thus, small changes in \(x\) (\(|x-9| < \delta\)) lead to small changes in \(\sqrt{x}\).

This attribute makes continuous functions predictable and easier to handle when working with limits.

Limit finding methods

Several methods exist for finding limits in calculus, providing tools to tackle complex problems:

• **Substitution**: The simplest, where you substitute the point into the function to find the limit.
• **Factoring**: Especially useful in polynomial expressions, it simplifies limits by canceling terms.
• **Rationalization**: Effective with roots, converting them to more manageable forms using conjugates.
• **Squeeze theorem**: If a function is bounded by two limits that converge to the same point, it also approaches this point.

When calculating limits using \(\delta\) like \(|x-2| < \frac{0.01}{12}\), the context often decides the method. Knowing which method applies best comes with practice and understanding, using logical steps to approach the form simpler enough to deal with delta-epsilon definitions or simpler inequality manipulations.