Problem 27 - Calculus: Early Transcendentals

Question

Compute the following limits. $$ \lim _{x \rightarrow 0^{+}} \frac{\sqrt{x+1}+1}{\sqrt{x+1}-1} $$

Step-by-Step Solution

Answer

The limit is \( +\infty \).

Step 1: Simplify the Expression using Conjugate

To simplify the expression, multiply the numerator and denominator by the conjugate of the denominator. This helps eliminate the square root in the denominator.The expression is \( \frac{\sqrt{x+1}+1}{\sqrt{x+1}-1} \).Multiply by \( \frac{\sqrt{x+1} + 1}{\sqrt{x+1} + 1} \):\[\frac{(\sqrt{x+1}+1)(\sqrt{x+1}+1)}{(\sqrt{x+1}-1)(\sqrt{x+1}+1)} = \frac{(\sqrt{x+1}+1)^2}{(\sqrt{x+1})^2 - 1^2}\]

Step 2: Apply the Difference of Squares Formula

The denominator \((\sqrt{x+1})^2 - 1^2\) is a difference of squares and can be simplified as follows:\[\frac{(\sqrt{x+1}+1)^2}{(x+1) - 1} = \frac{\sqrt{x+1}^2 + 2\sqrt{x+1} + 1}{x}\]

Step 3: Substitute Limits and Simplify

Now substitute the limit as \( x \to 0^+ \) into the simplified expression.The expression becomes:\[\frac{x + 2\sqrt{x+1} + 1}{x}\]Simplifying:\[\frac{(x + 1) + 1 + 2\sqrt{x+1}}{x} = \frac{x + 2\sqrt{x+1} + 1}{x}\]Cancelling gives:\[\frac{x}{x} + \frac{2\sqrt{x+1}}{x} + \frac{1}{x}\]

Step 4: Evaluate the Limit Components

As \( x \rightarrow 0^+ \):1. \( \frac{x}{x} \to 1 \)2. \( \frac{2\sqrt{x+1}}{x} \to \frac{2}{0^+} = \text{approaches infinity} \)3. \( \frac{1}{x} \to \frac{1}{0^+} = \text{approaches infinity} \)Thus, the limit is determined primarily by the terms that approach infinity.

Step 5: Conclude the Evaluation

The dominant terms \( \frac{2\sqrt{x+1}}{x} \) and \( \frac{1}{x} \) approach infinity as \( x \to 0^+ \), indicating that the expression itself becomes undefined in that neighborhood, diverging to \( +\infty \).

Key Concepts

Conjugate Multiplication

When handling limits that involve square roots, conjugate multiplication becomes a powerful tool. The idea is to multiply the numerator and denominator by the conjugate of the denominator. This process helps simplify expressions by removing the square root from the denominator.

Consider \[ \frac{\sqrt{x+1}+1}{\sqrt{x+1}-1} \].
To simplify it, multiply by the conjugate of the denominator, which is \(\sqrt{x+1} + 1\).
This yields:

• The numerator becomes \((\sqrt{x+1}+1)^2\).
• The denominator simplifies by the formula \((\sqrt{x+1})^2 - 1^2\).

This step is crucial as it not only makes the calculation more straightforward but also exposes the hidden complexity, making it manageable.

Difference of Squares

The difference of squares is a classic algebraic identity that frequently appears in limit calculations. It is generally expressed as \(a^2 - b^2 = (a-b)(a+b)\).
In this exercise, after applying conjugate multiplication, the denominator becomes:\[ (\sqrt{x+1})^2 - 1^2 = x+1-1 = x \].
This transformation is essential. It simplifies the expression considerably by reducing the complexity of the denominator. Expressing it as:

• The numerator: \((\sqrt{x+1}+1)^2\).
• The denominator: \(x\).

This simplification allows us to easily substitute the limit value in the expression later on. Ensuring calculations are not only correct but also manageable.

Indeterminate Forms

In calculus, some expressions may result in indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) when evaluated at certain limits. These forms do not immediately give a clear answer and require further manipulation.

Within this exercise scenario, the transformed expression becomes:\[ \frac{x + 2\sqrt{x+1} + 1}{x} \].
When you start cancelling terms or substituting the limit as \(x \to 0^+\),

• \(\frac{x}{x} \to 1\)
• \(\frac{2\sqrt{x+1}}{x}\) becomes an indeterminate form as it approaches infinity.
• \(\frac{1}{x}\) directly approaches infinity.

The indeterminate nature signifies the need for deeper evaluation to understand the "dominating" behaviour of the expression around the limit point.

Infinity in Limits

The concept of approaching infinity in limits is pivotal. When analyzing limits, determining various component behaviors is essential, especially when expressions diverge.

In the original problem, as \(x \to 0^+\), key components of the simplified equation approach infinity. Specifically:

• The term \(\frac{2\sqrt{x+1}}{x}\) tends towards infinity.
• Also, \(\frac{1}{x}\) goes toward infinity.

This indicates terms like these can heavily influence the result, primarily when they are dominant, resulting in the overall expression diverging to \(+\infty\).

Recognizing such behaviors enables a better grasp of the expression's behavior around the limit, supporting clearer conclusions about the nature of the divergence.