Problem 20 - Single Variable Calculus

Question

evaluate the limit using l'Hôpital's Rule if appropriate. $$ \lim _{x \rightarrow 1} \frac{a^{\ln x}-x}{\ln x} $$

Step-by-Step Solution

Answer

The limit is 0 after applying l'Hôpital's Rule and simplifying the expression: \[ \lim _{x \rightarrow 1} \frac{a^{\ln x} - x}{\ln x} = 0 \]

Step 1: Check for indeterminate form

Evaluate the limit as x approaches 1: \( \frac{a^{\ln(1)}-1}{\ln(1)} = \frac{a^0-1}{0} = \frac{1 - 1}{0} = \frac{0}{0} \) The limit is in the indeterminate form 0/0, so we can use l'Hôpital's Rule.

Step 2: Apply l'Hôpital's Rule

To apply l'Hôpital's Rule, we need to find the derivatives of both the numerator and denominator with respect to x. Let's do that: f(x) = \(a^{\ln(x)} - x\) f'(x) = \(\frac{d}{dx}(a^{\ln(x)} - x) = \frac{d}{dx}(a^{\ln(x)})-\frac{d}{dx}(x)\) Now, we apply the chain rule to find the derivative of \(a^{\ln(x)}\): \(\frac{d}{dx}(a^{\ln(x)}) = a^{\ln(x)} \cdot \frac{d}{dx}({\ln(x)}) = a^{\ln(x)} \cdot \frac{1}{x}\) So, f'(x) = \(\frac{a^{\ln(x)}}{x} - 1\) g(x) = \(\ln(x)\) g'(x) = \(\frac{d}{dx}(\ln(x)) = \frac{1}{x}\) Now we have: \(\lim _{x \rightarrow 1} \frac{a^{\ln x}-x}{\ln x} = \lim _{x \rightarrow 1} \frac{f'(x)}{g'(x)} = \lim _{x \rightarrow 1} \frac{\frac{a^{\ln(x)}}{x} - 1}{\frac{1}{x}}\)

Step 3: Simplify and evaluate the limit

Let's simplify \( \lim _{x \rightarrow 1} \frac{\frac{a^{\ln(x)}}{x} - 1}{\frac{1}{x}}\) by multiplying both numerator and denominator by x: \(\lim _{x \rightarrow 1} \frac{a^{\ln x}-x}{\ln x} = \lim _{x \rightarrow 1} \frac{a^{\ln x} - x}{1}\) Now, substitute x = 1 into the simplified expression: \(\lim _{x \rightarrow 1} \frac{a^{\ln x} - x}{1} = \frac{a^{\ln(1)} - 1}{1} = \frac{1-1}{1} = 0\) The limit is 0.

Key Concepts

Indeterminate Forms

In calculus, indeterminate forms present a challenge when trying to evaluate limits. They typically arise when substitution into a limit yields an expression like 0/0 or ∞/∞, among others. These forms are called 'indeterminate' because they do not provide immediate information about the behavior of a function as it approaches the limit.

Consider the exercise \( \lim _{x \rightarrow 1} \frac{a^{\ln x}-x}{\ln x} \). Upon first inspection, this may appear to be straightforward, but as you plug in x = 1, both the numerator and the denominator become zero. This gives us the classic indeterminate form of 0/0. It is in these cases that l'Hôpital's Rule becomes invaluable, as it provides a method to resolve the indeterminate nature and find the actual limit. The key is to differentiate both the numerator and the denominator until an evaluatable limit is achieved.

Limits in Calculus

Limits are a foundational concept in calculus, allowing mathematicians to understand the behavior of functions as they approach a certain point. They can indicate the value that a function approaches as the input approaches some value. Limits help in defining derivatives, integrals, and continuity of functions.

In the provided exercise, the task is to evaluate \( \lim _{x \rightarrow 1} \frac{a^{\ln x}-x}{\ln x} \) using l'Hôpital's Rule. The difficulty in directly evaluating this limit lies in the indeterminate form encountered. Limits can be evaluated from the left (lim_x→c^-) or right (lim_x→c⁺), but for continuous functions, and if both one-sided limits exist and are equal, the two-sided limit (lim_x→c) exists and is equal to the one-sided limits. Functions that are not continuous at a point may still have a limit, but additional techniques, such as l'Hôpital's Rule, may be necessary to find it.

Derivative

The derivative is a measure of how a function changes as its input changes. It's a concept from calculus that provides the rate at which a function is changing at any given point, often thought of as the slope of the tangent line to the function's graph at that point.

In the context of limits and l'Hôpital's Rule, derivatives are used to resolve indeterminate forms by applying differentiation to the numerator and the denominator of the function separately. This process often simplifies the expression enough to determine the limit. In the exercise, derivatives of \( f(x) = a^{\ln(x)} - x \) and \( g(x) = \ln(x) \) are calculated as \( f'(x) = \frac{a^{\ln(x)}}{x} - 1 \) and \( g'(x) = \frac{1}{x} \) respectively. Here, we see the application of the chain rule and standard differentiation techniques. After differentiation, we are left with a new ratio of derivatives which can be used to more easily find the sought limit.