Problem 133 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \lim _{x \rightarrow 1} \frac{x^{m}-1}{x^{n}-1}\left\\{\text { Ans. } \frac{m}{n}\right\\} $$

Step-by-Step Solution

Answer

\( \lim_{x \rightarrow 1} \frac{x^m-1}{x^n - 1} = \frac{m}{n} \)

Step 1: Identify indeterminate form

As x approaches 1, both the numerator and denominator approaches 0 when m and n are positive integers. So, we have an indeterminate form of the type \(\frac{0}{0}\).

Step 2: Apply L'Hopital's Rule

Since we have the indeterminate form \(\frac{0}{0}\), we can apply L'Hopital's rule which states that: \[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)} \] if the limit on the right-hand side exists. Before applying L'Hôpital's rule, let's find the derivatives of the numerator and denominator with respect to x.

Step 3: Calculate the derivatives

Find the derivative of \(x^m - 1\) with respect to x: \[ \frac{d}{dx} (x^m - 1) = mx^{m-1} \] Find the derivative of \(x^n - 1\) with respect to x: \[ \frac{d}{dx} (x^n - 1) = nx^{n-1} \]

Step 4: Apply L'Hopital's Rule again and find the limit

Now, we apply L'Hopital's rule to the function: \[ \lim_{x \rightarrow 1} \frac{x^m-1}{x^n - 1} = \lim_{x \rightarrow 1} \frac{mx^{m-1}}{nx^{n-1}} \] Since the right-hand side limit exists, we can proceed to evaluate the limit: \[ \lim_{x \rightarrow 1} \frac{mx^{m-1}}{nx^{n-1}} = \frac{m \cdot 1^{m-1}}{n \cdot 1^{n-1}} \] As \(1^{m-1} = 1\) and \(1^{n-1} = 1\), the limit simplifies to: \[ \frac{m}{n} \] Therefore, \[ \lim_{x \rightarrow 1} \frac{x^m-1}{x^n - 1} = \frac{m}{n} \]

Key Concepts

Indeterminate Forms

When dealing with calculus, particularly limits, you may encounter expressions that do not immediately yield a clear answer. These are known as indeterminate forms. A common type occurs when both the numerator and the denominator of a fraction approach zero as the variable approaches a certain value. This specific type is called a \( \frac{0}{0} \) indeterminate form. Why are they important?- Indeterminate forms occur frequently in calculus and indicate the need for a deeper dive into the limit to determine the actual behavior of the function.- Noticing an indeterminate form suggests that straightforward substitution won't work, and you must apply other techniques such as algebraic manipulation or, in many cases, L'Hopital's Rule.In our example, as \( x \) approaches 1, both \( x^m - 1 \) and \( x^n - 1 \) tend to 0, leading to a \( \frac{0}{0} \) form. This signals that L'Hopital's Rule is a viable method for evaluating the limit.

Derivatives

Derivatives are a core concept in calculus used to understand the rate at which a function is changing at any given point. For a function \( f(x) \), the derivative \( f'(x) \) represents the gradient or slope of the function at \( x \).Here's how derivatives help:- They provide the tool for translating a problem involving an indeterminate form into something more manageable.- In the context of limits and L'Hopital's Rule, calculating the derivative can change an indeterminate form into a form where the limit can be directly evaluated.For our limit problem, we need the derivatives of both \( x^m - 1 \) and \( x^n - 1 \):

• The derivative of \( x^m - 1 \) is \( mx^{m-1} \).
• The derivative of \( x^n - 1 \) is \( nx^{n-1} \).

By differentiating the numerator and the denominator, we apply L'Hopital's Rule, transforming the problem into a simpler limit that evaluates to \( \frac{m}{n} \).

Limits

In calculus, limits are used to describe the behavior of a function as it approaches a certain point. Understanding limits is crucial because they form the foundation of defining both derivatives and integrals.What makes limits useful?- They enable us to handle points where functions may not be explicitly defined, such as points of discontinuity or indeterminate forms.- Limits provide a method to examine the behavior of complex functions without necessarily knowing their explicit values.In our example, calculating \( \lim_{x \to 1} \frac{x^m-1}{x^n-1} \) involves recognizing the indeterminate form and applying L'Hopital's Rule. Once we use the derivatives, we find the limit simplifies nicely to \( \frac{m}{n} \), because \( x=1 \) causes each \( x^{m-1} \) and \( x^{n-1} \) to equal 1.In summary, limits help in bridging the gap between undefined expressions and the understanding of functions around those points.