Problem 20 - Calculus for Biology and Medicine

Question

Find the limits in Problems 1-60; not all limits require use of l'Hôpital's rule. $$ \lim _{x \rightarrow 0} \frac{e^{x}-1}{3 e^{2 x}-x} $$

Step-by-Step Solution

Answer

The limit is \( \frac{1}{5} \).

Step 1: Recognize the Indeterminate Form

First, substitute \( x = 0 \) directly into the function \( \frac{e^{x}-1}{3e^{2x}-x} \). This gives \( \frac{e^0 - 1}{3e^{0} - 0} = \frac{1-1}{3 \cdot 1 - 0} = \frac{0}{3} = 0 \).However, re-evaluate by observing how the numerator and denominator behave as \( x \to 0 \) to find that they both approach zero, indicating that this is an indeterminate form \( \frac{0}{0} \).

Step 2: Apply l'Hôpital's Rule

Since we identified the limit as an indeterminate form \( \frac{0}{0} \), we can apply l'Hôpital's Rule, which states that \( \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \) when the original limit is \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).

Step 3: Differentiate the Numerator and Denominator

Calculate the derivative of the numerator: \( \frac{d}{dx}(e^x - 1) = e^x \).Calculate the derivative of the denominator: \( \frac{d}{dx}(3e^{2x} - x) = 6e^{2x} - 1 \).

Step 4: Evaluate the New Limit

We now evaluate the new limit:\[ \lim_{x \to 0} \frac{e^x}{6e^{2x} - 1} \]Substitute \( x = 0 \) into this expression:\[ \frac{e^0}{6e^{0} - 1} = \frac{1}{6 \cdot 1 - 1} = \frac{1}{5} \].

Step 5: Final Conclusion: Limit Calculated

Thus, the limit \( \lim _{x \rightarrow 0} \frac{e^{x}-1}{3 e^{2 x}-x} \) is equal to \( \frac{1}{5} \).

Key Concepts

l'Hôpital's Rule

When dealing with some limits in calculus, you may encounter expressions that result in undefined or indeterminate forms. This is where **l'Hôpital's Rule** becomes handy! This rule is specifically designed to help in cases where substituting a limit value results in forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). It allows us to find the limit of a fraction by differentiating its numerator and denominator.

For the rule to apply, the original function must correspond to either of the two mentioned indeterminate forms. Once confirmed, here's what you generally do:

• Differentiate the numerator of the function.
• Differentiate the denominator of the function.
• Find the limit of the new fraction if the new limit doesn't result in another indeterminate form.

If after the first application, you still have an indeterminate form, l'Hôpital's rule can be applied repeatedly until a determinate form is reached.

Indeterminate Forms

In calculus, **indeterminate forms** are expressions that do not have a straightforward calculation when it comes to limits. Understanding these forms is crucial as they often show up in limits' problems.

There are several common indeterminate forms:

• \( \frac{0}{0} \)
• \( \frac{\infty}{\infty} \)
• \( 0 \cdot \infty \)
• \( \infty - \infty \)
• \( 0^0 \)
• \( \infty^0 \)
• \( 1^\infty \)

Each of these forms requires special techniques to evaluate limits, such as algebraic manipulation, applying l'Hôpital’s Rule, or using series expansions.

Identifying indeterminate forms in a problem is the first step to deciding the appropriate method to solve the limit.

Calculus Derivatives

**Calculus derivatives** are a fundamental part in understanding how functions change at any point. Derivatives give us the slope of the tangent line to the function at any particular point, which is a measure of the function's rate of change.

When applying l'Hôpital's rule, you often need to determine the derivatives of both the numerator and the denominator functions. This involves:

• Using derivative rules like the power rule, product rule, or chain rule, depending on the complexity of the functions involved.
• Calculating the derivative precisely for each part of the expression.
• Simplifying the resulting function, if possible, to evaluate the limit.

In the exercise above, we derived \( e^x - 1 \) and received \( e^x \), while deriving \( 3e^{2x} - x \) led to \( 6e^{2x} - 1 \). Afterwards, substituting \( x = 0 \) into these derived expressions enabled the proper calculation of the limit.