Problem 52 - Thomas Calculus

Question

Let \(\lim _{x \rightarrow 1} h(x)=5, \lim _{x \rightarrow 1} p(x)=1,\) and \(\lim _{x \rightarrow 1} r(x)=2\) Name the rules in Theorem 1 that are used to accomplish steps (a), (b), and (c) of the following calculation. $$ \lim _{x \rightarrow 1} \frac{\sqrt{5 h(x)}}{p(x)(4-r(x))}=\frac{\lim _{x \rightarrow 1} \sqrt{5 h(x)}}{\lim _{x \rightarrow 1}(p(x)(4-r(x)))}\quad\quad\quad\quad\quad\quad\text{(a)} \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad =\frac{\sqrt{\lim _{x \rightarrow 1} 5 h(x)}}{\left(\lim _{x \rightarrow 1} p(x)\right)\left(\lim _{x \rightarrow 1}(4-r(x))\right)}\quad\quad\quad\text{(b)} \quad\quad\quad \begin{aligned} &=\frac{\sqrt{5 \lim _{x \rightarrow 1} h(x)}}{\left(\lim _{x \rightarrow 1} p(x)\right)\left(\lim _{x \rightarrow 1} 4-\lim _{x \rightarrow 1} r(x)\right)}\quad\text{(c)} \\\ &=\frac{\sqrt{(5)(5)}}{(1)(4-2)}=\frac{5}{2} \end{aligned} $$

Step-by-Step Solution

Answer

The limit is \( \frac{5}{2} \).

Step 1: Separate the Limits

To separate the limit of the quotient, we use the Quotient Law for limits, which states that \( \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} \) provided \( \lim_{x \to a} g(x) eq 0 \). This allows us to write: \[ \lim_{x \rightarrow 1} \frac{\sqrt{5 h(x)}}{p(x)(4-r(x))} = \frac{\lim_{x \rightarrow 1} \sqrt{5 h(x)}}{\lim_{x \rightarrow 1}(p(x)(4-r(x)))}. \] This is step (a) in the given solution.

Step 2: Apply Limits to Composite Functions

Next, we apply the limit to the inner function within \( \sqrt{5h(x)} \) and the product \( p(x)(4-r(x)) \) separately using the Limit Law for compositions and products: \[ \frac{\sqrt{\lim_{x \rightarrow 1} 5 h(x)}}{\left(\lim_{x \rightarrow 1} p(x)\right)\left(\lim_{x \rightarrow 1}(4-r(x))\right)}. \] This simplifies the expression further, corresponding to step (b).

Step 3: Substitute Known Limits

Finally, substitute the known limits into the equation. For \( \sqrt{5h(x)} \), the Composite Limit Law applies, allowing: \[ \sqrt{5 \lim_{x \rightarrow 1} h(x)} = \sqrt{5 \times 5}. \] For the denominator, use the Product and Difference Limit Laws: \[ (\lim_{x \rightarrow 1} p(x)) \times (4 - \lim_{x \rightarrow 1} r(x)) = 1 \times (4 - 2). \] This corresponds to step (c). The expression simplifies to: \[ \frac{\sqrt{25}}{2}. \]

Step 4: Final Result: Simplify the Expression

Compute the final expression: \( \frac{\sqrt{25}}{2} = \frac{5}{2} \). This is the limit of the original expression as \( x \to 1 \).

Key Concepts

Limit Laws

Limit laws are fundamental rules in calculus that help us evaluate limits more easily. These laws allow us to manipulate complex expressions by breaking them down into simpler parts whose limits we already know. Here's a breakdown of some key limit laws:

• Constant Law: If you have a constant value, the limit is just that constant value.
• Sum/Difference Law: The limit of a sum (or difference) of functions is the sum (or difference) of their limits if both limits exist.
• Product Law: The limit of a product of functions is the product of their limits, assuming that both limits exist.
• Power Law: If a function is raised to a power, the limit can be taken as the function's limit raised to that power.

These laws make it simpler to deal with the behavior of functions as they approach a certain point or infinity. By utilizing these laws strategically, we can find the limits of even the most complex functions.

Quotient Law for Limits

The Quotient Law is particularly useful when dealing with ratios of functions. It states:\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} \]provided that the limit of the denominator is not zero, i.e., \(\lim_{x \to a} g(x) eq 0\). This law allows us to separate the limits when dealing with fractions, simplifying our computations.

• First, verify that the limit of the denominator is non-zero to ensure the quotient is defined.
• Next, apply the law to evaluate the limits of the numerator and the denominator independently.

This law is essential to solving complex limits involving fractional expressions, as demonstrated in Step 1 of the original exercise.

Composite Limit Law

The Composite Limit Law deals with functions nested within other functions. It allows us to evaluate limits by applying the limit laws to the inner and outer functions separately. Formally, if \(g(x)\) is continuous at \(x = a\) and \(f(x)\) is another function, then:\[ \lim_{x \to a} f(g(x)) = f(\lim_{x \to a} g(x)) \]provided \(\lim_{x \to a} g(x)\) exists.

• Identify the inner and outer functions for which the composite limit needs to be calculated.
• First, evaluate the limit of the inner function.
• Next, apply the outer function to this limit.

This principle helps simplify nested expressions, allowing for a step-by-step approach in solving limits of combined functions effectively.

Difference Limit Law

The Difference Limit Law is applied when subtracting one function from another. This law states that:\[ \lim_{x \to a} (f(x) - g(x)) = \lim_{x \to a} f(x) - \lim_{x \to a} g(x) \]provided both individual limits exist. It's helpful when dealing with expressions involving subtraction, as in the denominator of the original exercise.

• Ensure both functions have limits at the point of interest.
• Subtract the limits calculated for each part of the expression.

Using this law, we can evaluate the behavior of differences as \(x\) approaches the point in question, simplifying evaluations of more complex limit problems.