Problem 149 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \lim _{x \rightarrow 0} \frac{e^{-x}-1}{x}\\{\text { Ans. }-1\\} $$

Step-by-Step Solution

Answer

Using L'Hôpital's Rule, we first verify that the limit is in indeterminate form 0/0. Then, differentiate the numerator and the denominator: \( \frac{d}{dx}(e^{-x} - 1) = -e^{-x} \) and \( \frac{d}{dx}(x) = 1 \). The new ratio becomes: \( \frac{-e^{-x}}{1} \). Lastly, calculate the limit as x approaches 0: \( \lim_{x \rightarrow 0} \frac{-e^{-x}}{1} = \frac{-e^0}{1} = \frac{-1}{1} \). Therefore, the answer is \( \lim_{x \rightarrow 0} \frac{e^{-x} - 1}{x} = -1 \).

Step 1: Verify that L'Hôpital's Rule can be applied

First, let's check if we can apply L'Hôpital's Rule. We need to show that the limit of the numerator and the denominator are both 0 or both infinity. As x approaches 0: \( \lim_{x \rightarrow 0} e^{-x} = e^0 = 1 \) \( \lim_{x \rightarrow 0} -1 = -1 \) \( \lim_{x \rightarrow 0} x = 0 \) Thus, the limit is in the indeterminate form 0/0, and we can apply L'Hôpital's Rule.

Step 2: Differentiate the numerator and the denominator

Now let's differentiate the numerator and the denominator with respect to x: \( \frac{d}{dx}(e^{-x} - 1) = -e^{-x} \) \( \frac{d}{dx}(x) = 1 \) So, the new ratio becomes: \( \frac{-e^{-x}}{1} \)

Step 3: Calculate the limit using the new ratio

Now, we need to calculate the limit of the new ratio as x approaches 0: \( \lim_{x \rightarrow 0} \frac{-e^{-x}}{1} = \frac{-e^0}{1} = \frac{-1}{1} \) The limit of the given function as x approaches 0 is -1. Therefore, the answer is: \( \lim_{x \rightarrow 0} \frac{e^{-x} - 1}{x} = -1 \)

Key Concepts

Indeterminate Forms

When tackling limits in calculus, we often come across expressions that are not immediately clear in terms of their behavior as the input approaches a certain value. These expressions are referred to as indeterminate forms. Examples of indeterminate forms include 0/0, \(\infty / \infty\), 0 \cdot \infty, \infty - \infty, 0^0, \infty^0, and 1^\infty. These forms are called indeterminate because they do not lead to a definitive limit and require further mathematical techniques to evaluate.

An indeterminate form like 0/0, which appeared in the exercise of finding the limit of \(\frac{e^{-x}-1}{x}\) as x approaches 0, does not imply the limit does not exist. Rather, it signals that we must do more work to find the limit, such as using L'Hôpital's Rule, factoring, rationalizing, or finding a common denominator to simplify the expression and reveal the limit.

Limits in Calculus

One of the cornerstone concepts in calculus is the limit. It is used to understand the behavior of functions as the input gets infinitely close to a particular point. The limit of a function as the variable approaches a specific value can tell us many things, such as the value that a function is approaching, the behavior of a function near that value, or even if a function can produce a certain value. In our textbook exercise, we are presented with the task of finding \(\lim_{x \rightarrow 0} \frac{e^{-x}-1}{x}\).

To find this limit, it's important to evaluate the behavior of the numerator and denominator separately. If both simultaneously approach zero or infinity, we have an indeterminate form; here L'Hôpital's Rule comes into play, allowing us to differentiate the numerator and denominator to potentially reveal a clear limit.

Differentiation

Differentiation is the process of finding the derivative of a function, which gives us the rate at which the function's value is changing at any given point. In essence, the derivative reflects the function's slope at a particular point. Within the context of L'Hôpital's Rule, we use differentiation to resolve indeterminate forms. By finding the derivatives of both the numerator and denominator separately, as seen in our exercise with the function \(\frac{e^{-x}-1}{x}\), we create a new function which can then be analyzed for its limit as x approaches a specific value.

In our original problem, after differentiating the numerator and denominator, we ended up with \(\frac{-e^{-x}}{1}\). The derivative simplifies the process of finding the limit, as we evaluate the resulting function's behavior when x approaches 0, leading us to the conclusive limit of -1.