Problem 11 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The limit is infinity as \( x \to 0^+ \).

Step 1: Identify the Form

Observe that as \( x \to 0^+ \), both the numerator \( \sqrt{x} \rightarrow 0 \) and the denominator \( \ln(x+1) \rightarrow 0 \). This results in an indeterminate form \( \frac{0}{0} \). Thus, l'Hôpital's Rule is applicable.

Step 2: Apply l'Hôpital's Rule

Since we have an indeterminate form, apply l'Hôpital's Rule which states that \( \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \), provided the right limit exists. Compute the derivatives: \( f(x) = \sqrt{x} \) has the derivative \( f'(x) = \frac{1}{2\sqrt{x}} \), and \( g(x) = \ln(x+1) \) has the derivative \( g'(x) = \frac{1}{x+1} \).

Step 3: Evaluate the New Limit

The limit becomes \( \lim_{x \to 0^+} \frac{\frac{1}{2\sqrt{x}}}{\frac{1}{x+1}} = \lim_{x \to 0^+} \frac{x+1}{2\sqrt{x}} \). Simplify this to \( \lim_{x \to 0^+} \frac{x+1}{2x^{1/2}} \).

Step 4: Simplify and Compute the Limit

Observe the expression \( \lim_{x \to 0^+} \frac{x+1}{2x^{1/2}} = \lim_{x \to 0^+} \left( \frac{1}{2}x^{-1/2} + \frac{1}{2x^{1/2}} \right) \). As \( x \to 0^+ \), the dominant term \( \frac{1}{2}x^{-1/2} \) tends towards infinity. Therefore, the limit is infinity.

Key Concepts

l'Hôpital's Rule

Understanding l'Hôpital's Rule is essential for tackling certain types of limit problems in calculus, especially those involving indeterminate forms. In essence, l'Hôpital's Rule provides a method for resolving limits that at first glance appear indeterminate, typically forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). The rule states that under appropriate conditions, the limit of the ratio of two functions can be found by taking the limit of the ratio of their derivatives:

\[\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\]provided that the limit on the right-hand side exists.

When to Use l'Hôpital's Rule?

When a limit evaluates to an indeterminate form like \( \frac{0}{0} \):

• Check if differentiating the numerator and denominator can resolve the indeterminacy.
• Ensure the functions are differentiable near the point of interest, except potentially at the point itself.

Example Through the Exercise

In the exercise given, \( \frac{\sqrt{x}}{\ln(x+1)} \), both the numerator and denominator approach zero as \( x \to 0^+ \). By applying l'Hôpital's Rule, we differentiate to find the derivatives \( f'(x) = \frac{1}{2\sqrt{x}} \) and \( g'(x) = \frac{1}{x+1} \), transforming the original indeterminate form into something more manageable that can lead to an accurate evaluation of the limit.

limits

Limits are a fundamental concept in calculus, describing the behavior of a function as its input approaches a particular value. Essentially, evaluating limits allows us to understand the trend of function values, whether they stabilize, shoot off to infinity, or oscillate. This is crucial when functions do not necessarily have defined values at particular points but still exhibit a predictable pattern as those points are approached.

Why Limits Matter

• They help in understanding the continuity and differentiability of functions.
• Limits are foundational in defining derivatives and integrals, key operations in calculus.
• They provide insights into the behavior of functions at boundary points or points of discontinuity.

Evaluating Limits

When solving limits, you might encounter forms that initially provide no clear direction, such as 0/0. This is where advanced techniques such as substitution, factoring, or the use of l'Hôpital's Rule come in. In our exercise, the limit \( \lim_{x \rightarrow 0^+} \frac{\sqrt{x}}{\ln (x+1)} \) highlights a case for l'Hôpital’s Rule due to the 0/0 form.

By re-expressing the limit through derivatives, we found a clearer path to the solution, ultimately leading to simplification and evaluation of the transformed limit expression.

indeterminate forms

Indeterminate forms occur when the direct substitution in a limit results in expressions like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), or other undefined expressions like \( 0 \times \infty \). These forms are particularly intriguing because they suggest that further analysis is required, often leading to solutions that are not immediately apparent.

Common Indeterminate Forms

• \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \) - can often be addressed using l'Hôpital's Rule.
• Forms such as \( 0^0 \), \( \infty - \infty \), and \( 1^{\infty} \) - often require different strategies such as algebraic manipulation or series expansion.

Resolving Indeterminate Forms

Proper handling of indeterminate forms involves recognizing the situation and selecting an appropriate strategy, such as:

• Applying l'Hôpital's Rule to differentiate the numerator and denominator.
• Rewriting expressions to facilitate easier evaluation, perhaps through factoring or algebraic rearrangement.
• Utilizing trigonometric identities, series, or even numeric approximation for more complex scenarios.

In the given exercise, recognizing the \( \frac{0}{0} \) form allowed the use of l'Hôpital’s Rule to effectively analyze the limit as \( x \to 0^+ \). This involved calculating derivatives to reformulate and find a valid solution.