Problem 21 - Calculus Early Transcendentals

Question

Evaluate the limit, if it exists. $$\lim _{t \rightarrow 9} \frac{9-t}{3-\sqrt{t}}$$

Step-by-Step Solution

Answer

The limit is 6.

Step 1: Plug in the Limit Value

First, substitute the value that \( t \) approaches, which is 9, into the function. This would give us \( \frac{9-9}{3-\sqrt{9}} = \frac{0}{0} \). This is an indeterminate form, so we need to manipulate the function to evaluate the limit.

Step 2: Rationalize the Denominator

To resolve the indeterminate form, multiply the numerator and the denominator by the conjugate of the denominator, \(3+\sqrt{t}\). This gives us \( \frac{(9-t)(3+\sqrt{t})}{(3-\sqrt{t})(3+\sqrt{t})} \). The denominator simplifies to \((3-\sqrt{t})(3+\sqrt{t}) = 9 - t\).

Step 3: Simplify the Expression

Notice that \(9-t\) appears in both the numerator and the denominator, so we can cancel it out. After canceling, the expression becomes \(3+\sqrt{t}\).

Step 4: Evaluate the Limit

Substitute \( t = 9 \) back into the simplified expression \(3+\sqrt{t}\). This evaluates to \(3+\sqrt{9} = 3+3 = 6\). Therefore, the limit is 6.

Key Concepts

Indeterminate Forms

In calculus, when evaluating limits, you may encounter expressions that yield certain results such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) upon direct substitution. These are known as **indeterminate forms**. They don't lead directly to a specific number, leaving the true limit value unclear without further manipulation.

The expression \( \lim_{t \to 9} \frac{9-t}{3-\sqrt{t}} \), when substituted directly with 9, results in the form \( \frac{0}{0} \). This means direct evaluation is not possible and you must use algebraic techniques such as factorization or rationalization to resolve the indeterminate form. This makes it essential to **reanalyze the expression** to find an alternative way to evaluate the limit.

Rationalization

Rationalization is a technique primarily used to eliminate radicals, especially in denominators. It is a powerful method for handling indeterminate forms like \( \frac{0}{0} \).

In the given exercise, the denominator \(3-\sqrt{t}\) poses a problem when \(t\) approaches 9. By multiplying the numerator and the denominator by the conjugate \(3+\sqrt{t}\), you effectively remove the radical from the denominator, transforming it into \( 9 - t \), which can further simplify the expression.

This step is crucial because it helps transform the expression into a form that's easily manageable and enables the simplification needed to solve the limit problem accurately.

Evaluating Limits

After simplifying a function of a limit problem, the next step is to **evaluate the limit** by substitution—provided the expression is now in a form where this is possible.

From the exercise, following rationalization and cancellation, the complex function \( \frac{(9-t)(3+\sqrt{t})}{9-t} \) was reduced to the simpler \(3+\sqrt{t}\). Now, by substituting \( t = 9 \) into this expression, it results in: \( 3 + \sqrt{9} = 6 \).

This step shows how algebraic manipulation can guide the evaluator to obtain a precise limit result, offering clarity where initially there appeared to be none due to the indeterminate form.