Problem 10 - Calculus for Biology and Medicine

Question

Find the limits in Problems 1-60; not all limits require use of l'Hôpital's rule. $$ \lim _{x \rightarrow \infty} \sqrt{x}-\sqrt{x+3} $$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Consider the Expression

The given limit is \( \lim_{x \rightarrow \infty} (\sqrt{x} - \sqrt{x+3}) \). Rewrite this expression for clarity in manipulation.

Step 2: Multiply and Divide by Conjugate

Multiply and divide the expression by the conjugate, which is \( \sqrt{x} + \sqrt{x+3} \). So the expression becomes \((\sqrt{x} - \sqrt{x+3}) \cdot \frac{\sqrt{x} + \sqrt{x+3}}{\sqrt{x} + \sqrt{x+3}}\).

Step 3: Simplify the Expression

When multiplying conjugates, you'll have \((\sqrt{x})^2 - (\sqrt{x+3})^2 = x - (x+3) = -3\) in the numerator. Thus, the expression simplifies to \(-\frac{3}{\sqrt{x} + \sqrt{x+3}}\).

Step 4: Evaluate the Limit

Evaluate \( \lim_{x \to \infty} -\frac{3}{\sqrt{x} + \sqrt{x+3}} \). As \(x\) approaches infinity, \(\sqrt{x}\) and \(\sqrt{x+3}\) both approach infinity, making the denominator very large. Hence, the entire expression approaches zero.

Key Concepts

Conjugate method

The conjugate method is a highly effective technique in calculus, especially when dealing with limits involving square roots. It helps to simplify expressions into a more manageable form. This technique involves multiplying and dividing the expression by its conjugate. For instance, if you have an expression such as \( \sqrt{x} - \sqrt{x+3} \), its conjugate would be \( \sqrt{x} + \sqrt{x+3} \).

By applying the conjugate method, you effectively employ the difference of squares formula. This is because the expression \((a - b)(a + b)\) simplifies to \(a^2 - b^2\). In our original problem, using the conjugate results in the expression \(-\frac{3}{\sqrt{x} + \sqrt{x+3}}\), after simplifying \( (\sqrt{x})^2 - (\sqrt{x+3})^2 = x - (x+3) = -3\).

This simplification removes the square roots, simplifying the evaluation of limits as \(x\) approaches infinity. After applying the conjugate, the problem becomes analyzing the limit of a rational function, making it much easier to solve.

Infinity in calculus

Infinity is a concept that appears frequently in calculus, especially in limits. When we say \(x\) approaches infinity, we imply that \(x\) is growing without bound. It helps in evaluating behaviors of functions as the input values are extremely large.

As seen in the problem, both \(\sqrt{x}\) and \(\sqrt{x+3}\) approach infinity. As these terms grow, their difference is manipulated to find the function's limit. This is why the initial expression may seem undetermined, but simplifying with tools like conjugates reveals the true limiting behavior.

Understanding infinity is crucial for solving limits. It represents an unrestricted growth rather than a specific number. This helps in analyzing how functions behave in extreme cases, enabling predictions about their growth, decline, or convergence.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus for finding limits, particularly those involving indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). However, it's not always necessary. Sometimes alternate methods, such as the conjugate method, are more straightforward. Nevertheless, it's vital to understand how this rule works.

When you encounter a limit of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), L'Hôpital's Rule allows you to differentiate the numerator and the denominator separately, and then reevaluate the limit. This can simplify complex expressions and clarify the limiting behavior.

Although L'Hôpital's Rule wasn't required for the solution of our specific problem, recognizing when it's applicable can save time and reduce errors in other complex limit evaluations.