Problem 29 - Calculus: Early Transcendentals

Question

Evaluate the limit, if it exists. \( \displaystyle \lim_{t \to 0}\left( \frac{1}{t \sqrt{1 + t}} - \frac{1}{t} \right)\)

Step-by-Step Solution

Answer

The limit is \(-\frac{1}{2}\).

Step 1: Identify the Problem Type

We need to evaluate the limit \( \lim_{t \to 0}\left( \frac{1}{t \sqrt{1 + t}} - \frac{1}{t} \right)\). This expression is of the indeterminate form \( \frac{0}{0} \) as \( t \to 0 \). Therefore, we'll either simplify or manipulate it to evaluate the limit.

Step 2: Combine the Fractions

Combine the two fractions under a common denominator: \[\frac{1}{t \sqrt{1 + t}} - \frac{1}{t} = \frac{1 - \sqrt{1 + t}}{t \cdot \sqrt{1 + t}}\]

Step 3: Simplify the Numerator using Conjugates

Multiply and divide by the conjugate of \(1 - \sqrt{1 + t}\), which is \(1 + \sqrt{1 + t}\) to simplify the numerator:\[\frac{(1 - \sqrt{1 + t})(1 + \sqrt{1 + t})}{t \cdot (\sqrt{1 + t})(1 + \sqrt{1 + t})}\]Simplifying the numerator gives:\[1^2 - (\sqrt{1+t})^2 = 1 - (1+t) = -t\]so the expression becomes:\[\frac{-t}{t(\sqrt{1 + t})(1 + \sqrt{1 + t})}\]

Step 4: Cancel Common Factors

Cancel \(t\) from the numerator and denominator:\[\frac{-1}{(\sqrt{1 + t})(1 + \sqrt{1 + t})}\]

Step 5: Evaluate the Limit

Now evaluate the limit as \( t \to 0 \):\[\lim_{t \to 0} \frac{-1}{(\sqrt{1 + t})(1 + \sqrt{1 + t})} = \frac{-1}{(\sqrt{1 + 0})(1 + \sqrt{1 + 0})} = \frac{-1}{1 \cdot 2} = -\frac{1}{2}\]

Step 6: Conclusion: Final Answer

The limit exists and its value is \( -\frac{1}{2} \).

Key Concepts

Indeterminate Forms

Indeterminate forms in calculus arise when the direct substitution of a value into an expression results in an undefined or ambiguous result, often categorized by forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms are important to recognize because they signal that further analysis or manipulation of the expression is needed to evaluate a limit effectively. In our given exercise, as \( t \to 0 \), the expression \( \frac{1}{t \sqrt{1 + t}} - \frac{1}{t} \) initially results in the indeterminate form \( \frac{0}{0} \). This situation requires us to apply specific techniques, like algebraic simplification or special limit theorems, to resolve the ambiguity and find a defined limit. Recognizing and addressing indeterminate forms is critical in calculus to ensure proper evaluation and understanding of limits.

Limit Simplification Techniques

Simplifying an expression to evaluate a limit often involves algebraic manipulations that make a seemingly complex limit more straightforward to solve. The step-by-step solution to our exercise showcases these techniques very well. When we face an indeterminate form such as \( \frac{0}{0} \), combining terms over a common denominator or rationalizing the expression can often clear the confusion and lead us towards a solution. In the exercise, the fractions were first combined under a common denominator, transforming the original expression to \( \frac{1 - \sqrt{1 + t}}{t \cdot \sqrt{1 + t}} \). This step aims to shape the problem into a more manageable form. Once simplified, cancelling common terms in the numerator and the denominator often reveals the limit in a way that allows direct evaluation, as we saw with the cancellation of \( t \) in this exercise. These simplification techniques are invaluable tools for solving limits in calculus.

Conjugate Multiplication in Limits

Conjugate multiplication is an effective technique often used to simplify limits involving square roots or irrational expressions. The concept involves multiplying an expression by a "conjugate," which for a difference involving square roots is the expression with the opposite operation in the middle. For our exercise, this meant taking the expression \( 1 - \sqrt{1 + t} \) and multiplying by its conjugate \( 1 + \sqrt{1 + t} \), both in the numerator and the denominator. This process is beneficial because it exploits the identity \((a - b)(a + b) = a^2 - b^2\), leading to a simpler form for the expression. In our case, using the conjugate multiplication transformed the numerator into \( 1^2 - (\sqrt{1 + t})^2 = 1 - (1+t) = -t \), which considerably simplified the process and allowed for subsequent cancellation of terms. Using conjugates helps eliminate roots, making the algebra easier to manage, and is especially useful when dealing with indeterminate forms.