Question

\(\lim _{\mathrm{h} \rightarrow 5}\left[\left\\{(2 \mathrm{~h}+5)^{(5 / 2)}-(15)^{(5 / 2)}\right\\} /\left(\mathrm{h}^{3}-125\right)\right]=?\) (a) \(\sqrt{5}\) (b) \(\sqrt{125}\) (c) \(\sqrt{15}\) (d) \((15)^{(5 / 2)}\)

Short answer

(c) \(\sqrt{15}\)

Step-by-step solution

1. Rewrite the expression in terms of the given limit

First, also recall that we typically use the notation \(\lim_{x \rightarrow a}\) to denote the limit as x approaches a value a. In our case, we'd have the given limit as \(\lim_{h \rightarrow 5}\). Now, rewrite the expression in terms of the given limit: \[ \lim_{h \rightarrow 5} \frac{(2h +5)^{\frac{5}{2}} - (15)^{\frac{5}{2}}}{h^3 - 125} \]

2. Identify the indeterminate form

Next, we need to identify the indeterminate form of this limit by substituting the given value of h, which is 5: \[ \frac{(2(5) + 5)^{\frac{5}{2}} - (15)^{\frac{5}{2}}}{(5)^3 - 125} \] We can simplify this as: \[ \frac{(15)^{\frac{5}{2}} - (15)^{\frac{5}{2}}}{(125) - 125} \] Which looks like: \[ \frac{0}{0} \] This is an indeterminate form.

3. Apply L'Hopital's Rule

Since we have the indeterminate form \(\frac{0}{0}\), we can apply L'Hôpital's Rule for finding the limit. L'Hôpital's Rule states that if \(\lim_{x \rightarrow a} \dfrac{f(x)}{g(x)} = \dfrac{0}{0}\) or \(\dfrac{\infty}{\infty}\), then the limit can be found by differentiating both the numerator and denominator: \[ \lim_{x \rightarrow a} \dfrac{f(x)}{g(x)} = \lim_{x \rightarrow a} \dfrac{f'(x)}{g'(x)} \] Let's find the derivatives of the numerator and denominator: \(f(h) = (2h + 5)^{\frac{5}{2}} - (15)^{\frac{5}{2}}\) \(f'(h) = \frac{5}{2}(2h + 5)^{\frac{3}{2}}(2)\) \(g(h) = h^3 - 125\) \(g'(h) = 3h^2\) Now, using L'Hôpital's Rule, we plug these derived functions back in to find the limit: \[ \lim_{h \rightarrow 5} \frac{f'(h)}{g'(h)} = \lim_{h \rightarrow 5} \frac{\frac{5}{2}(2h + 5)^{\frac{3}{2}}(2)}{3h^2} \]

4. Solve the limit

Finally, plug in the value h = 5: \[ \lim_{h \rightarrow 5} \frac{\frac{5}{2}(2(5) + 5)^{\frac{3}{2}}(2)}{3(5)^2} = \frac{\frac{5}{2}(15)^{\frac{3}{2}}(2)}{3(25)} \] Simplify the expression: \[ \frac{5(15)^{\frac{3}{2}}}{3(25)} \] This reduces to: \[ \frac{15^{\frac{5}{2}}}{\sqrt{15}} \] Thus, the answer is \(\boxed{\text{(c)} \sqrt{15}}\).

Key concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful method in calculus for solving limits that result in indeterminate forms. When you evaluate a limit and find that it leads to expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), this rule can be applied to simplify and find the limit more easily.

What the rule tells us is that when you have such an indeterminate form, you can take the derivative of the numerator and the derivative of the denominator separately, and then find the limit of the resulting fraction.

For example, if \( \lim_{x \rightarrow a} \frac{f(x)}{g(x)} \) turns into \( \frac{0}{0} \), you can find the derivative of \( f(x) \) and \( g(x) \), denoted as \( f'(x) \) and \( g'(x) \). Then, compute \( \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)} \). This new limit will often be much simpler to evaluate.

• Keep in mind this rule only applies to these specific indeterminate forms.
• You might have to apply the rule more than once if the resulting limit is still indeterminate.

Indeterminate forms

Indeterminate forms are expressions that arise when evaluating limits and they don't lead easily to a definitive result. These forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), and others like \( 0 \cdot \infty \), \( \infty - \infty \), and \( 1^\infty \). The most common scenarios occur with fractions where substitution leads to \( \frac{0}{0} \) or similar outcomes.

The presence of an indeterminate form suggests that direct substitution in a limit won't work and alternative strategies, such as L'Hôpital's Rule or algebraic manipulation, must be used.

Indeterminate forms are crucial in calculus because they signal that a problem needs more sophisticated methods to solve. They are like red flags that tell us we need to dig deeper. In our example, when substituting \( h = 5 \), both the numerator and denominator turn to zero, giving us the \( \frac{0}{0} \) indeterminate form.

Differentiation

Differentiation is a key technique in calculus used to find the derivative of a function. The derivative measures how a function changes as its input changes, essentially capturing the rate of change in the function.

When applying L'Hôpital's Rule, differentiation is necessary. You need to differentiate both the numerator and the denominator of the function. In our example, we took the derivative of \((2h + 5)^{\frac{5}{2}}\) and \(h^3 - 125\).

• Product and Chain Rule: Often, you will employ these rules to differentiate complex expressions.
• Simplification: After differentiation, make sure to simplify expressions before evaluating the limit again.

Understanding differentiation is crucial in problem-solving within calculus, as it allows you to handle functions analytically, applying rules and algebra to break down complex equations into more manageable parts.

Calculus problem solving

Solving calculus problems often involves a series of strategic steps to arrive at a solution. These typically begin with understanding the problem, identifying possible hindrances like indeterminate forms, and then selecting appropriate methods to tackle them.

In our limit problem, we first rewrote the expression to make it easier to handle. We then substituted to check for indeterminate forms and, finding \( \frac{0}{0} \), opted to use L'Hôpital's Rule. Performing the differentiation accurately and simplifying gave us the final solution.

• Plan Your Approach: It's beneficial to outline how you will tackle a problem. Choose the right tool, whether it's substitution, factoring, or L'Hôpital's Rule.
• Check Your Work: After solving, rethink the steps taken to ensure no mistakes occurred.
• Practice: Regular practice is essential to mastering calculus problem-solving techniques. Ensures familiarity with various approaches and their applications.

These strategies help in tackling a variety of calculus problems efficiently while reinforcing your understanding of fundamental concepts.