Problem 6 - Calculus for Biology and Medicine

Question

Find the limits in Problems 1-60; not all limits require use of l'Hôpital's rule. $$ \lim _{x \rightarrow 0} \frac{\sqrt{2 x+4}-2}{x} $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Indeterminate Form

First, we need to evaluate the expression inside the limit as \(x\) approaches 0. Substituting \(x = 0\) gives us \(\frac{\sqrt{2 \times 0 + 4} - 2}{0}\), which simplifies to \(\frac{2 - 2}{0} = \frac{0}{0}\). This is an indeterminate form 0/0, which means we need to find another method to evaluate the limit.

Step 2: Rationalize the Numerator

To resolve the indeterminate form, we rationalize the numerator. Multiply the numerator and the denominator by the conjugate of the numerator, \(\sqrt{2x + 4} + 2\). This gives us:\[\frac{(\sqrt{2x + 4} - 2)(\sqrt{2x + 4} + 2)}{x(\sqrt{2x + 4} + 2)}\]

Step 3: Simplify the Expression

By using the difference of squares formula, the numerator becomes:\[(\sqrt{2x + 4})^2 - 2^2 = 2x + 4 - 4 = 2x\]So, the expression now simplifies to:\[\frac{2x}{x(\sqrt{2x + 4} + 2)}\]

Step 4: Cancel Common Terms

Now, we can cancel \(x\) in the numerator and the denominator since \(x eq 0\):\[\frac{2x}{x(\sqrt{2x + 4} + 2)} = \frac{2}{\sqrt{2x + 4} + 2}\]

Step 5: Evaluate the Limit as x Approaches 0

Substitute \(x = 0\) into the simplified expression:\[\lim_{x \to 0} \frac{2}{\sqrt{2 \times 0 + 4} + 2} = \frac{2}{2 + 2} = \frac{2}{4} = \frac{1}{2}\]

Key Concepts

l'Hôpital's Rule

When you encounter a calculus problem involving limits, particularly when the direct substitution leads to an indeterminate form like \(\frac{0}{0}\), l'Hôpital's Rule can be a very useful tool. This rule states that if \(\lim_{{x \to a}} f(x)/g(x)\) results in \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then:\[\lim_{{x \to a}} \frac{f(x)}{g(x)} = \lim_{{x \to a}} \frac{f'(x)}{g'(x)}\]This means we differentiate the numerator and the denominator separately. However, it's crucial to ensure that the conditions for using the rule are met and the limit exists or else the method is invalid. l'Hôpital's Rule is particularly helpful in simplifying complex rational functions where direct evaluation is not possible.

Indeterminate Forms

An indeterminate form arises when the limit of a function cannot be directly computed due to ambiguities in the mathematical expression, often seen as \(\frac{0}{0}\), \(\infty - \infty\), or other similar forms. These forms can be tricky as they do not immediately provide insight into the limit.

• In the given exercise, substituting \(x = 0\) resulted in \(\frac{0}{0}\), indicating an indeterminate form.
• To resolve this, we must apply alternative methods such as l'Hôpital's Rule, algebraic manipulation, or trigonometric identities to simplify and evaluate the limit accurately.

Understanding when a function results in an indeterminate form tells us that further algebraic or analytical manipulation is necessary to find the limit.

Rationalizing the Numerator

Rationalizing the numerator is a technique used to eliminate square roots in the numerator, making it easier to simplify expressions. This involves multiplying both the numerator and the denominator by the conjugate of the numerator. The conjugate is essentially the same expression but with the sign between terms reversed.In this exercise, the expression \(\sqrt{2x + 4} - 2\) in the numerator is problematic when \(x\) is substituted as zero. By multiplying by its conjugate, \(\sqrt{2x + 4} + 2\), we use the difference of squares to eliminate the square root:- \((\sqrt{2x + 4} - 2)(\sqrt{2x + 4} + 2) = (\sqrt{2x + 4})^2 - 2^2 = 2x + 4 - 4 = 2x\)This process simplifies the expression, allowing us to cancel terms and efficiently evaluate the limit.

Difference of Squares

The difference of squares is a powerful algebraic identity used to simplify expressions and solve equations. It follows the principle that:\[(a - b)(a + b) = a^2 - b^2\]This identity is often applied when dealing with square roots, as seen in this problem:

• By rationalizing the numerator with its conjugate, we directly use the difference of squares: \((\sqrt{2x + 4} - 2)(\sqrt{2x + 4} + 2) = (\sqrt{2x + 4})^2 - 2^2\).
• This results in the elimination of the square roots, simplifying to \(2x\), which is much easier to handle.

By simplifying expressions using the difference of squares, we can effectively resolve complex problems involving limits that present indeterminate forms.