Problem 29 - Calculus Early Transcendentals

Question

Evaluate the limit, if it exists. $$\lim _{t \rightarrow 0}\left(\frac{1}{t \sqrt{1+t}}-\frac{1}{t}\right)$$

Step-by-Step Solution

Answer

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

Step 1: Rewrite the Expression

Begin by rewriting the expression with a common denominator. The original expression is \( \frac{1}{t \sqrt{1+t}} - \frac{1}{t} \). Rewrite it as \( \frac{1 - \sqrt{1+t}}{t \sqrt{1+t}} \), where \( t \) is already a part of the denominator.

Step 2: Rationalize the Numerator

Multiply the numerator and the denominator by the conjugate of the numerator, \( 1 + \sqrt{1+t} \). This gives \( \frac{(1 - \sqrt{1+t})(1 + \sqrt{1+t})}{t \sqrt{1+t} (1 + \sqrt{1+t})} \).

Step 3: Simplify the Expression

The numerator simplifies using the difference of squares: \( (1 - \sqrt{1+t})(1 + \sqrt{1+t}) = 1 - (1 + t) = -t \). Thus, the expression becomes \( \frac{-t}{t \sqrt{1+t}(1 + \sqrt{1+t})} \).

Step 4: Cancel Common Factors

Cancel \( t \) in the numerator and denominator: \( \frac{-1}{\sqrt{1+t}(1 + \sqrt{1+t})} \).

Step 5: Evaluate the Limit

Evaluate the limit as \( t \rightarrow 0 \), substituting \( t = 0 \) into \( \frac{-1}{\sqrt{1+t}(1 + \sqrt{1+t})} \): \( \frac{-1}{\sqrt{1} \times (1 + 1)} = \frac{-1}{2} \).

Key Concepts

Rationalizing Expressions

Rationalizing expressions is a technique used to eliminate radicals from the denominator of a fraction. The method involves multiplying the numerator and the denominator by the conjugate of the term containing the radical. The conjugate is simply the same terms with the opposite sign in between them.
\(
\)In the given problem, the expression \( \frac{1 - \sqrt{1+t}}{t \sqrt{1+t}} \) is multiplied by the conjugate of the numerator, \( 1 + \sqrt{1+t} \). This changes the expression to:

• Numerator: \( (1 - \sqrt{1+t})(1 + \sqrt{1+t}) \)
• Denominator: \( t \sqrt{1+t} \times (1 + \sqrt{1+t}) \)

This process doesn't change the value of the expression because we simply multiply by 1 in the form of conjugate/conjugate.

Difference of Squares

When dealing with conjugates, the difference of squares formula comes in handy. The formula is \( a^2 - b^2 = (a-b)(a+b) \). This helps to easily simplify expressions where conjugates are being multiplied together.
\(
\)In the context of the given limit problem, \( (1 - \sqrt{1+t})(1 + \sqrt{1+t}) \) simplifies using the difference of squares to:

• \( 1^2 - (\sqrt{1+t})^2 = 1 - (1 + t) \)
• This results in \( 1 - 1 - t = -t \).

The difference of squares simplifies the complex expression into something much more manageable.

Common Denominator

Finding a common denominator allows you to combine two fractions into a single expression. This can be an essential step when simplifying complex fractional expressions.
\(
\)In the original problem, you begin with two separate fractions: \( \frac{1}{t \sqrt{1+t}} \) and \( \frac{1}{t} \).
By using a common denominator, the expression \( \frac{1 - \sqrt{1+t}}{t \sqrt{1+t}} \) emerges. This makes subsequent steps of rationalizing straightforward because it brings the fractions under a cohesive term.

Cancel Common Factors

The final step in simplifying rational expressions often involves canceling out common factors in the numerator and the denominator. This step reduces the expression to its simplest form, making it easier to evaluate.
\(
\)In the example, once the numerator \( -t \) is obtained, the \( t \) in both the numerator and denominator can be canceled out. This cancellation is legitimate since \( t eq 0 \)—as \( t \rightarrow 0 \), we use other algebraic techniques first to avoid division by zero.
\(
\)Cancellation leaves the expression as \( \frac{-1}{\sqrt{1+t}(1+\sqrt{1+t})} \). With a simpler fraction, we can proceed to evaluate the limit as \( t \rightarrow 0 \).