Problem 6 - Calculus for Biology and Medicine

Question

Use l'Hospital's rule to find the limits. $$ \lim _{x \rightarrow 0} \frac{3-\sqrt{2 x+9}}{2 x} $$

Step-by-Step Solution

Answer

The limit is \(-\frac{1}{6}\).

Step 1: Verify Indeterminate Form

First, substitute \( x = 0 \) into the expression \( \frac{3-\sqrt{2x+9}}{2x} \). This gives \( \frac{3-\sqrt{9}}{0} = \frac{0}{0} \). Since the limit evaluates to \( \frac{0}{0} \), it is in an indeterminate form suitable for l'Hospital's Rule.

Step 2: Differentiate the Numerator

Differentiate the numerator, \( 3-\sqrt{2x+9} \) with respect to \( x \). The derivative of \( 3 \) is \( 0 \) and the derivative of \( -\sqrt{2x+9} \) is \( \frac{-2}{2\sqrt{2x+9}} \) using the chain rule. So the derivative of the numerator is \( \frac{-2}{2\sqrt{2x+9}} \).

Step 3: Differentiate the Denominator

Differentiate the denominator, which is \( 2x \), with respect to \( x \). The derivative of \( 2x \) is simply \( 2 \).

Step 4: Apply l'Hospital's Rule

Apply l'Hospital's Rule to the limit: \[ \lim _{x \to 0} \frac{3-\sqrt{2x+9}}{2x} = \lim _{x \to 0} \frac{\frac{-2}{2\sqrt{2x+9}}}{2}. \]Simplifying gives \[ \lim _{x \to 0} \frac{-1}{2\sqrt{2x+9}}. \]

Step 5: Evaluate the Limit

Now, substitute \( x = 0 \) into the expression from Step 4:\[ \frac{-1}{2\sqrt{9}} = \frac{-1}{6}. \]

Key Concepts

Understanding Indeterminate Forms

When we encounter limits in calculus, sometimes substituting directly gives us expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These expressions are known as *indeterminate forms*. They do not have a straightforward value, making them tricky to evaluate directly.
In our exercise, substituting \( x = 0 \) into the limit \( \lim _{x \rightarrow 0} \frac{3-\sqrt{2x+9}}{2x} \) gives \( \frac{3-\sqrt{9}}{0} = \frac{0}{0} \). This result isn't defined mathematically and signals an indeterminate form.

• Indeterminate forms indicate a need for manipulation to evaluate the limit.
• Common indeterminate forms include \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \).
• Techniques like l'Hospital's Rule can resolve these forms by differentiating the numerator and denominator.

By identifying the indeterminate form, we recognize it's suitable to apply l'Hospital's Rule to find the limit.

Using the Chain Rule for Differentiation

Differentiation is a key process in calculus used to measure how a function changes as its input changes. Sometimes, functions are composed of other functions, making differentiation a bit more involved.
The *chain rule* is a tool that allows us to differentiate these composite functions efficiently. It states that if you have a composite function \( f(g(x)) \), its derivative is \( f'(g(x)) \cdot g'(x) \).
In our problem, we differentiate the function \( 3 - \sqrt{2x + 9} \):

• The derivative of the constant \( 3 \) is \( 0 \).
• To differentiate \( -\sqrt{2x + 9} \), you notice it's a composite function. Think of it as \( -(2x + 9)^{1/2} \).
  • Use \( \frac{d}{dx} (u)^{1/2} = \frac{1}{2}u^{-1/2} \cdot \frac{du}{dx} \) where \( u = 2x + 9 \).
  • We find \( u' = 2 \), leading to \(-\frac{1}{2} (2x+9)^{-1/2} \times 2\).

As a result, the derivative of the function is \( \frac{-2}{2\sqrt{2x + 9}} \), which is crucial for applying l'Hospital's Rule later.

Limit Evaluation Strategies

Evaluating limits is a crucial skill in calculus, especially when traditional substitution results in an indeterminate form.
One of the most effective techniques for evaluating limits of indeterminate forms is l'Hospital's Rule, which allows us to compute a limit by differentiating both the numerator and denominator independently.

• First, confirm the expression is in the form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
• Differentiation simplifies the expressions, often removing the indeterminacy.
• After applying l'Hospital's Rule in our problem:
  • We differentiate the numerator, resulting in \( \frac{-2}{2\sqrt{2x+9}} \).
  • The denominator derivative is \( 2 \).
  Substitute these into the limit expression:
  
• Easiest form to evaluate: \[ \lim _{x \to 0} \frac{-2}{2 \times 2\sqrt{2x+9}} \]

Substituting \( x = 0 \) into the simplified form gives a clear value: \( \frac{-1}{6} \). Therefore, the limit is successfully evaluated, and the indeterminate form resolved.