Problem 9 - Calculus: Early Transcendentals

Question

Compute the following limits. $$ \lim _{t \rightarrow 0}\left(t+\frac{1}{t}\right)\left((4-t)^{3 / 2}-8\right) $$

Step-by-Step Solution

Answer

The limit is -6.

Step 1: Recognize the Indeterminate Form

As we substitute \( t = 0 \) directly into the expression \( \left(t + \frac{1}{t}\right)((4-t)^{3/2} - 8) \), we encounter an indeterminate form \( \frac{0}{0} \). This suggests the use of L'Hôpital's Rule or algebraic manipulation to resolve the limit.

Step 2: Algebraic Expansion

Notice that for small values of \( t \), \((4-t)^{3/2}\) can be expanded using a binomial series. Specifically:\[(4-t)^{3/2} = 8 - 3\sqrt{4}t + \cdots = 8 - 6t + \cdots\]Thus, the expression becomes:\[\left(t + \frac{1}{t}\right)(-6t + \cdots)\]

Step 3: Simplify the Expression

Focus on simplifying:\[t\left(-6t\right) + \frac{1}{t}\left(-6t\right) = -6t^2 - 6\]As \( t \rightarrow 0 \), the first term \(-6t^2\) approaches zero, and we are left with the constant term \(-6\).

Step 4: Find the Limit

Given the simplification, the remaining expression does not depend on \( t \), hence:\[\lim_{t \rightarrow 0}(-6) = -6\]

Step 5: Conclude the Solution

Thus, evaluating the limit after simplification, we find that the limit is a constant value, \(-6\), as \( t \) approaches zero.

Key Concepts

Indeterminate Forms

In calculus, indeterminate forms arise when applying basic limit laws yield ambiguous or undefined results. When you substitute a specific value into a limit expression, you might encounter forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These indicate that the limit cannot be directly determined using simple arithmetic, and require further analysis or algebraic manipulation.

For example, in the given exercise, substituting \( t = 0 \) results in an indeterminate form \( \frac{0}{0} \). Understanding this allows us to conclude that special techniques like L'Hôpital's Rule or series expansions may be necessary to evaluate the limit accurately.

Indeterminate forms signal that something more subtle must be happening in the behavior of the functions as they approach the limit. Recognizing these forms is the first step in solving more complex limit problems. This knowledge helps decide on techniques like L'Hôpital's Rule or algebraic manipulation to find the true behavior of the limit around the point of interest.

L'Hôpital's Rule

L'Hôpital's rule is a powerful tool in calculus for finding the limits of indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). The rule states that if the limit of a function results in an indeterminate form, you can take the derivative of the numerator and the derivative of the denominator, then find the limit of that new function.

The rule can be symbolically stated as:

• If \( \lim_{x \to a} f(x) = 0 \ \text{and} \ \lim_{x \to a} g(x) = 0 \)
• or \( \lim_{x \to a} f(x) = \pm \infty \ \text{and} \ \lim_{x \to a} g(x) = \pm \infty \),

Then,\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]provided the limit on the right-hand side exists. This is especially useful when direct substitution in a limit calculation yields \( \frac{0}{0} \) as seen in our original exercise.

However, sometimes L'Hôpital's Rule is not the most straightforward approach, particularly if additional simplifications like algebraic expansions or factoring are available. It's important to consider multiple angles while approaching limits, and L'Hôpital's Rule is just one powerful option in the toolkit.

Binomial Series Expansion

The binomial series expansion is a method to expand expressions involving powers of binomials for small perturbations from a known value. It is especially useful in approximating expressions when a variable is close to zero, facilitating the evaluation of limits that involve these expressions.

For a general expression like \( (1 + x)^n \), the binomial series expansion can be written as:

• \((1 + x)^n \approx 1 + nx + \frac{n(n-1)}{2!}x^2 + \cdots\)

In the context of the given exercise, we use this expansion to simplify \((4-t)^{3/2}\), where \( t \) is small. The expansion around \( t = 0 \) gives us:

• \((4-t)^{3/2} \approx 8 - 6t + \cdots\)

This simplification turns a complex expression into a more manageable polynomial form.

The choice to use a binomial series is strategic in that it simplifies calculations and provides a clearer view of the function's behavior as \( t \to 0 \). By using the series, you transform an indeterminate product involving a complicated root expression into simpler terms that more clearly reflect how each piece of the expression contributes to the overall limit.