Problem 5 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Evaluate the limits. a. \(\lim _{x \rightarrow 5} 3 x^{2}-15 x\) b. \(\lim _{x \rightarrow 0} 3 x\) c. \(\lim _{x \rightarrow 50} \pi\) d. \(\lim _{x \rightarrow-2} 3 x^{2}-15 x\) e. \(\lim _{x \rightarrow-\pi} x\) f. \(\lim _{x \rightarrow \pi} 50\) g. \(\lim _{x \rightarrow-5} \frac{x-1}{x-1}\) h. \(\lim _{x \rightarrow 1} \frac{x^{2}-2 x+1}{x-1}\) i. \(\quad \lim _{x \rightarrow 1} \frac{x^{4}-1}{x-1}\) j. \(\quad \lim _{x \rightarrow 4} \frac{x}{\sqrt{x+2 x^{2}}}\) k. \(\lim _{x \rightarrow 1} \sqrt{\frac{x-2}{x^{3}+1}}\) 1\. \(\lim _{x \rightarrow 1} \sqrt{\frac{x-3}{x^{3}-27}}\)

Step-by-Step Solution

Answer

a. 0, b. 0, c. \(\pi\), d. 42, e. \(-\pi\), f. 50, g. 1, h. 0, i. 4, j. \(\frac{2}{3}\), k. undefined, l. \(\frac{1}{\sqrt{13}}\).

Step 1: Evaluate Limit (a)

To find \(\lim _{x \rightarrow 5} 3x^{2} - 15x\), we can directly substitute \(x = 5\) into the expression. \[3(5)^2 - 15(5) = 3(25) - 75 = 75 - 75 = 0\] So, the limit as \(x\) approaches 5 of \(3x^{2} - 15x\) is 0.

Step 2: Evaluate Limit (b)

The limit \(\lim _{x \rightarrow 0} 3x\) can be found by substituting \(x = 0\) into the expression: \[3(0) = 0\]. So the limit is 0.

Step 3: Evaluate Limit (c)

For \(\lim _{x \rightarrow 50} \pi\), since \(\pi\) is a constant, the limit is simply \(\pi\).

Step 4: Evaluate Limit (d)

To solve \(\lim _{x \rightarrow -2} 3x^{2} - 15x\), substitute \(-2\) into the expression: \[3(-2)^2 - 15(-2) = 3(4) + 30 = 12 + 30 = 42\] Thus, the limit is 42.

Step 5: Evaluate Limit (e)

Since this is a direct substitution and \(x\) is approaching \(-\pi\) for the limit \(\lim _{x \rightarrow -\pi} x\), the limit equals \(-\pi\).

Step 6: Evaluate Limit (f)

For \(\lim _{x \rightarrow \pi} 50\), since 50 is a constant, the limit is simply 50.

Step 7: Evaluate Limit (g)

The expression \(\frac{x-1}{x-1}\) simplifies to 1, except where \(x = 1\). For \(x = -5\), since \(x-1\) does not approach 0 in the denominator, the limit is 1.

Step 8: Evaluate Limit (h)

The limit \(\lim _{x \rightarrow 1} \frac{x^{2}-2x+1}{x-1}\) is indeterminate since it results in \(\frac{0}{0}\). Factoring the numerator, we have \(\frac{(x-1)^2}{x-1} = x-1\). The limit then becomes \(\lim _{x \rightarrow 1} (x-1) = 0\).

Step 9: Evaluate Limit (i)

\(\lim _{x \rightarrow 1} \frac{x^4-1}{x-1}\) also gives \(\frac{0}{0}\), so factor the numerator: \(\frac{(x-1)(x^3+x^2+x+1)}{x-1}\). Simplify to \(x^3+x^2+x+1\) and substitute \(x = 1\): \[1^3 + 1^2 + 1 + 1 = 4\].

Step 10: Evaluate Limit (j)

Substitute \(x = 4\) into \(\frac{x}{\sqrt{x+2x^2}}\), we simplify by combining terms under the square root: \[\frac{4}{\sqrt{4+2(4)^2}} = \frac{4}{\sqrt{4+32}} = \frac{4}{\sqrt{36}} = \frac{4}{6} = \frac{2}{3}\].

Step 11: Evaluate Limit (k)

The limit \(\lim _{x \rightarrow 1} \sqrt{\frac{x-2}{x^3+1}}\) is undefined since substituting gives \(\frac{-1}{2}\), causing the square root to be undefined for real numbers.

Step 12: Evaluate Limit (l)

To find \(\lim _{x \rightarrow 1} \sqrt{\frac{x-3}{x^3 - 27}}\), attempting direct substitution yields \(\sqrt{\frac{-2}{-26}} = \sqrt{\frac{1}{13}}\), defining the limit as \(\frac{1}{\sqrt{13}}\).

Key Concepts

Evaluating Limits

In calculus, evaluating limits is a fundamental concept that helps us understand the behavior of functions as they approach a certain point. To evaluate a limit, we are trying to find out what value a function approaches as the input (or variable) approaches a certain value. This concept is crucial in defining derivatives and integrals, which are the building blocks of calculus.

When evaluating limits, it's important to consider both the value from the positive and negative sides of the number we are approaching. Limits can sometimes be straightforward but can also be complex, especially when involving indeterminate forms such as \(\frac{0}{0}\) or \infty - \infty\.

Here's a quick recap:

• A limit gives an idea of how a function behaves as it gets closer to a particular point.
• The process is essential in understanding discontinuities, infinite behaviors, and oscillatory functions.
• Limits can be evaluated using various techniques, such as direct substitution, factoring, rationalizing, or using L'Hôpital's Rule for indeterminate forms.

Direct Substitution Method

The direct substitution method is one of the simplest techniques used to find limits. When using this method, you simply replace the variable in the expression with the value it approaches. This is often the first step when evaluating limits because it's easy and quick.

Direct substitution works smoothly when:

• The function is continuous at the point you are evaluating; this means that there's no break, hole, or vertical asymptote at that point.
• The substitution does not result in an indeterminate form such as \(\frac{0}{0}\).

For example, in the exercise, evaluating limits like \(\lim_{x \rightarrow 0} 3x\) involves direct substitution by simply plugging in \ x = 0 \, resulting in \[3(0) = 0\].

However, direct substitution is not applicable in situations of indeterminate forms, and other algebraic manipulations or rules may be needed to evaluate the limit correctly. It's a good initial step to quickly check functions where direct evaluation might just work without additional calculations.

Indeterminate Forms

Indeterminate forms arise when evaluating limits that initially seem undefined, such as \(\frac{0}{0}\), \infty - \infty\, or \(\left(1^{\infty}\right)\). These forms indicate that the limit requires more work to determine its value. They reveal that at first glance, substituting the number gives us a meaningless result that doesn't convey the actual behavior of the function around that point.

The common indeterminate forms include:

• \(\frac{0}{0}\)
• \(\frac{\infty}{\infty}\)
• \(\infty - \infty\)
• \(0^{0}\)

When you encounter these forms, you often need to apply additional techniques. These may include algebraic manipulation like factoring, canceling common terms, or applying more advanced methods like L'Hôpital's Rule. For instance, in the exercise, expression \(\lim_{x \rightarrow 1} \frac{x^2 - 2x + 1}{x - 1}\) gives us a \(\frac{0}{0}\) form. Here, factoring helps simplify the expression to \(x - 1\), and its limit as \(x\) approaches 1 is zero.