Question

Use analytical methods to evaluate the following limits. $$\lim _{x \rightarrow 1} \frac{x \ln x-x+1}{x \ln ^{2} x}$$

Short answer

Answer: The limit of the given function as x approaches 1 is 1.

Step-by-step solution

Check if the limit is an indeterminate form.

As x approaches 1, both the numerator and denominator approach 0. To show this, plug in x=1 into the expression: $$\frac{1 \ln 1 - 1 + 1}{1 \ln^2 1} = \frac{0}{0}$$ Since we get the indeterminate form 0/0, we need to use L'Hôpital's rule to find the limit.

Apply L'Hôpital's rule.

L'Hôpital's rule states that if the limit of the ratio of two functions' derivatives exists and is finite, it's equal to the limit of the original functions' ratio. So, we need to find the derivatives of the numerator and the denominator with respect to x. Derivative of the numerator (\(f(x)\)): $$f'(x) = \frac{d}{dx}\left(x\ln x - x + 1\right) = \ln x + 1 - 1 = \ln x$$ Derivative of the denominator (\(g(x)\)): $$g'(x) = \frac{d}{dx}\left(x\ln^2 x\right) = \ln x + \ln^2 x$$ Now, find the limit of the ratio of the derivatives: $$\lim_{x \rightarrow 1} \frac{f'(x)}{g'(x)} = \lim_{x \rightarrow 1} \frac{\ln x}{\ln x + \ln^2 x}$$

Evaluate the limit of the ratio of derivatives.

As x approaches 1, both the numerator and the denominator of the new expression approach 0. However, since \(\ln x\) is a common factor, we can simplify the expression and find the limit directly: $$\lim_{x \rightarrow 1} \frac{\ln x}{\ln x + \ln^2 x} = \lim_{x \rightarrow 1} \frac{\ln x}{\ln x(1 + \ln x)} = \lim_{x \rightarrow 1} \frac{1}{1 + \ln x}$$ Now, plug in x=1 into the simplified expression: $$\lim_{x \rightarrow 1} \frac{1}{1 + \ln x} = \frac{1}{1 + \ln(1)} = \frac{1}{1} = 1$$ The limit of the given function as x approaches 1 is 1.

Key concepts

Limits

A limit in calculus helps us determine the value that a function approaches as the input approaches a particular point. This concept is crucial when dealing with continuous functions or trying to solve problems involving infinity or very small numbers. For example, if we have a function \( f(x) \), finding \( \lim_{x \to a} f(x) \) means identifying what value \( f(x) \) gets closer to as \( x \) approaches \( a \). The process of finding limits is essential in calculus, especially as it paves the way to understanding continuity and the behavior of functions at boundaries or discontinuities.

In our exercise, we have a limit where \( x \) approaches 1. Evaluating this helps us understand how the entire function behaves around this point, even if the function isn't defined at precisely \( x = 1 \). This is particularly useful in scenarios where direct substitution leads to uncertainties like \( \frac{0}{0} \).

L'Hôpital's Rule

L'Hôpital's Rule is a handy tool for finding limits that result in indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When direct substitution in a limit problem yields such forms, L'Hôpital's Rule allows us to differentiate the numerator and denominator separately and re-evaluate the limit.

For example, for a limit \( \lim_{x \to a} \frac{f(x)}{g(x)} \) with either \( 0/0 \) or \( \infty/\infty \), L'Hôpital's Rule suggests replacing it with \( \lim_{x \to a} \frac{f'(x)}{g'(x)} \), provided these derivatives exist. It's crucial to confirm that differentiation is simplifying the problem. In our exercise, after applying L'Hôpital's Rule, the limit simplifies, resulting in an easily computable value, allowing us to overcome the initial indeterminate form.

Derivative

A derivative represents the rate at which a function is changing at any point. It is fundamental in calculus for understanding behavior and finding slopes of curves. To compute the derivative of a function \( f(x) \), you analyze how \( f(x) \) changes as \( x \) changes by an infinitesimally small amount.

In terms of the exercise, the derivative of the numerator, \( x \ln x - x + 1 \), is \( \ln x \), and that of the denominator, \( x \ln^2 x \), is \( \ln x + \ln^2 x \). Calculating these derivatives lets us apply L'Hôpital's Rule effectively, converting a complex limit into something more manageable, eventually allowing us to solve the limit problem. Understanding derivatives is pivotal because it gives insight into much of calculus, including optimization and curve sketching.

Indeterminate Forms

In calculus, indeterminate forms appear when evaluating limits of ratios results in undefined expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These indicate uncertainty in calculation and can prevent straightforward determination of a limit.

In our exercise, substituting \( x = 1 \) directly into the given limit leads to the indeterminate form \( \frac{0}{0} \). Indeterminate forms signal the need for tools like L'Hôpital's Rule to re-evaluate using derivatives rather than straightforward algebraic substitution.

Resolving indeterminate forms is key in calculus because it ensures limits are computed accurately, unravelling true function behavior near specific points. This methodical approach helps in not just theoretical calculus classes but also practical applications, ensuring students and professionals alike can compute outcomes as \( x \) nears critical numbers.