Limit Of (e^x - 1) / (2x) As X->0: Step-by-Step Solution

Nov 18, 2025 by Alex Johnson 57 views

Finding the Limit of (e^x - 1) / (2x) as x Approaches 0

Are you wrestling with limits in calculus? Let's break down a classic problem: finding the limit of the function (e^x - 1) / (2x) as x gets closer and closer to 0. This is a fundamental concept in calculus, and understanding it will pave the way for tackling more complex problems. We'll explore different approaches and provide a step-by-step solution to make the process crystal clear.

Understanding Limits

Before diving into the problem, it’s crucial to understand what a limit actually means. In simple terms, a limit tells us what value a function approaches as its input (in this case, x) gets closer and closer to a specific value. It’s not necessarily the value of the function at that point, but rather the value it's tending towards.

In our problem, we want to find out what happens to the expression (e^x - 1) / (2x) as x gets infinitesimally close to 0. We can't simply plug in 0 directly, because that would result in division by zero, which is undefined. This is where the concept of limits comes into play. To properly grasp limits, think of it like this: we're not asking what the function is at x = 0, but where it's going as x gets extremely close to 0.

This understanding is key because many functions behave differently as they approach a point than they do at the point itself. The limit allows us to analyze the behavior of a function in these critical zones, which is essential for various applications in calculus and beyond, such as finding derivatives, integrals, and analyzing the continuity of functions. The formal definition of a limit involves the concept of arbitrarily small distances (epsilon-delta definition), but for our purposes, the intuitive understanding of approaching a value is sufficient. So, as we delve into solving this specific limit, keep in mind that we are exploring the function's behavior in the immediate vicinity of x = 0, not exactly at x = 0.

Methods to Solve Limits

There are several techniques we can use to solve limits, and the best approach often depends on the specific problem. Here are a few common methods:

Direct Substitution: This is the simplest method, where you plug in the value that x is approaching into the function. However, this only works if the function is continuous at that point and doesn't result in an undefined expression (like division by zero).
Factoring and Simplifying: If direct substitution leads to an indeterminate form (like 0/0), you might be able to factor the numerator and denominator and cancel out common factors. This can often simplify the expression and allow you to use direct substitution.
L'Hôpital's Rule: This powerful rule applies when you have an indeterminate form of the type 0/0 or ∞/∞. It states that the limit of the quotient of two functions is equal to the limit of the quotient of their derivatives. This is a very useful tool for many limit problems.
Using Standard Limits: Some limits are so common that they're considered "standard limits." Knowing these can help you solve more complex problems. For example, the limit of sin(x)/x as x approaches 0 is a classic standard limit that equals 1.
Series Expansion: For some functions, especially those involving trigonometric or exponential terms, you can use their series expansions (like Taylor or Maclaurin series) to approximate the function near a specific point. This can often simplify the limit calculation.

In our case, direct substitution of x = 0 into (e^x - 1) / (2x) results in (e^0 - 1) / (2 * 0) = (1 - 1) / 0 = 0/0, which is an indeterminate form. This tells us that we need to use a different method. Factoring and simplifying doesn't seem applicable here, and we don't have a standard limit that directly fits. So, L'Hôpital's Rule seems like a promising approach, as it's specifically designed for handling indeterminate forms like 0/0.

Applying L'Hôpital's Rule

Since direct substitution resulted in the indeterminate form 0/0, we can apply L'Hôpital's Rule. This rule states that if the limit of f(x)/g(x) as x approaches a is of the form 0/0 or ∞/∞, then:

lim (x→a) [f(x) / g(x)] = lim (x→a) [f'(x) / g'(x)]

where f'(x) and g'(x) are the derivatives of f(x) and g(x), respectively.

In our problem, f(x) = e^x - 1 and g(x) = 2x. Let's find their derivatives:

f'(x) = d/dx (e^x - 1) = e^x g'(x) = d/dx (2x) = 2

Now, we can apply L'Hôpital's Rule:

lim (x→0) [(e^x - 1) / (2x)] = lim (x→0) [e^x / 2]

Now, we can try direct substitution again. As x approaches 0, e^x approaches e^0, which is 1. Therefore:

lim (x→0) [e^x / 2] = 1 / 2

So, using L'Hôpital's Rule, we've found that the limit of (e^x - 1) / (2x) as x approaches 0 is 1/2. This method is powerful because it transforms a tricky indeterminate form into a simpler limit that we can evaluate directly. It's a cornerstone technique in calculus and is frequently used to solve a wide range of limit problems. Remember, the key is to identify when L'Hôpital's Rule is applicable (i.e., when you have an indeterminate form) and to correctly compute the derivatives of the numerator and denominator.

Alternative Approach: Using Series Expansion

While L'Hôpital's Rule provides a straightforward solution, let's explore another method to solve this limit: using the Maclaurin series expansion for e^x. The Maclaurin series is a special case of the Taylor series, which represents a function as an infinite sum of terms involving its derivatives at a single point (in this case, x = 0).

The Maclaurin series for e^x is given by:

e^x = 1 + x + (x^2 / 2!) + (x^3 / 3!) + (x^4 / 4!) + ...

where ! denotes the factorial function (e.g., 3! = 3 * 2 * 1 = 6). This series provides a way to approximate the value of e^x for values of x near 0. The more terms we include in the series, the better the approximation becomes. Now, let's substitute this expansion into our limit expression:

lim (x→0) [(e^x - 1) / (2x)] = lim (x→0) [(1 + x + (x^2 / 2!) + (x^3 / 3!) + ... - 1) / (2x)]

Notice that the '1' and '-1' terms cancel out, leaving us with:

lim (x→0) [(x + (x^2 / 2!) + (x^3 / 3!) + ...) / (2x)]

Now, we can factor out an 'x' from the numerator:

lim (x→0) [x(1 + (x / 2!) + (x^2 / 3!) + ...) / (2x)]

The 'x' in the numerator and denominator cancel each other out:

lim (x→0) [(1 + (x / 2!) + (x^2 / 3!) + ...) / 2]

Now, we can apply direct substitution. As x approaches 0, all the terms with 'x' in them will approach 0. This leaves us with:

(1 + 0 + 0 + ...) / 2 = 1 / 2

So, using the Maclaurin series expansion, we've arrived at the same answer: the limit of (e^x - 1) / (2x) as x approaches 0 is 1/2. This method demonstrates the power of series expansions in approximating functions and solving limit problems. It's particularly useful when dealing with functions that don't have simple algebraic forms or when other methods, like L'Hôpital's Rule, might be more cumbersome to apply.

Conclusion

We've successfully found the limit of (e^x - 1) / (2x) as x approaches 0 using two different methods: L'Hôpital's Rule and series expansion. Both approaches led us to the same answer: 1/2. This problem highlights the importance of understanding different limit-solving techniques and choosing the most appropriate one for the given situation.

Limits are a cornerstone of calculus, and mastering them is crucial for understanding more advanced concepts like derivatives, integrals, and continuity. By understanding various methods such as L'Hôpital's Rule and series expansion, you can tackle a wide range of limit problems with confidence. Remember, practice is key to mastering these techniques. Work through various examples and try different approaches to solidify your understanding.

For further exploration of limits and calculus, consider visiting resources like Khan Academy's Calculus section.