Problem 5 - Calculus for Biology and Medicine

Question

Use l'Hospital's rule to find the limits. $$ \lim _{x \rightarrow 0} \frac{\sqrt{2 x+4}-2}{x} $$

Step-by-Step Solution

Answer

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

Step 1: Identify Indeterminate Form

First, substitute the limit point, which is 0, into the expression \( \frac{\sqrt{2x+4}-2}{x} \). This gives us \( \frac{\sqrt{4} - 2}{0} \) which results in \( \frac{0}{0} \), an indeterminate form. This indicates that l'Hospital's Rule can be applied.

Step 2: Apply l'Hospital's Rule

To apply l'Hospital's Rule, differentiate the numerator and the denominator separately. The original function is \( \sqrt{2x+4} - 2 \). The derivative of the numerator, \( \sqrt{2x+4} - 2 \), is \( \frac{d}{dx}(\sqrt{2x+4}) = \frac{1}{2\sqrt{2x+4}} \times 2 = \frac{1}{\sqrt{2x+4}} \). The derivative of the denominator, \( x \), is 1.

Step 3: Rewrite the Limit Using Derivatives

Substitute these derivatives back into the limit expression using l'Hospital's rule: \( \lim _{x \rightarrow 0} \frac{\frac{1}{\sqrt{2x+4}}}{1} = \lim _{x \rightarrow 0} \frac{1}{\sqrt{2x+4}} \).

Step 4: Calculate the Limit

Now, substitute \( x = 0 \) into the expression \( \frac{1}{\sqrt{2x+4}} \). This results in \( \frac{1}{\sqrt{4}} = \frac{1}{2} \). Thus, the limit is \( \frac{1}{2} \).

Key Concepts

Indeterminate Forms

When working with limits, you might encounter expressions that do not initially yield a clear answer. These confusing forms are known as "indeterminate forms." Common examples include \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \). These forms are like placeholders that don't immediately tell us the limit's value but rather provide a clue that we might need another method to evaluate the limit.

For instance, in the given exercise \( \lim _{x \rightarrow 0} \frac{\sqrt{2x+4}-2}{x} \), plugging \(x = 0\) in directly leads to \( \frac{0}{0} \). This form is inconclusive and signals that we can potentially use l'Hospital's Rule to find the limit.

Remember, identifying indeterminate forms is crucial as it guides us on whether l'Hospital’s Rule is applicable. The next step involves differentiating the numerator and denominator to resolve this unclear division.

Limits

Limits in calculus help us understand how a function behaves as it approaches a specific point. They are fundamental for defining concepts like continuity, derivatives, and integrals.

In our exercise, we're interested in the behavior of the function \( f(x) = \frac{\sqrt{2x+4}-2}{x} \) as \( x \) approaches 0. By attempting to directly evaluate the function at this point, we encountered an indeterminate form. Thus, we look at the behavior of \( f(x) \) very close to \( x = 0 \) instead.

This process involves the formal method of calculating a limit, often represented as \( \lim_{x \to a} f(x) \), which means determining the value that \( f(x) \) approaches as \( x \) gets infinitely close to \( a \). In our case, finding this value gives insight into how the function behaves around this point and aids in solving the limit using l'Hospital's Rule.

Derivatives

Derivatives are tools in calculus used for understanding how a function changes at any point. They express the rate of change or slope of the function at a particular point.

Applying this to the exercise, we use the derivative to simplify the evaluation of the limit. When we find \( \frac{0}{0} \), l'Hospital's Rule allows us to handle this indeterminate form by differentiating the numerator and the denominator. We differentiate \( \sqrt{2x+4} - 2 \) with respect to \( x \), giving us \( \frac{1}{\sqrt{2x+4}} \). The derivative of \( x \) is straightforward: 1.

Insert these derivatives back into the limit expression to simplify it: \( \lim _{x \rightarrow 0} \frac{1}{\sqrt{2x+4}} \). This new limit can be evaluated directly, and finding the derivative was key to resolving the initial indeterminate form effectively.