Proving Limits Lim (x→0) (e^x - 1)/x = 1 And Lim (x→0) (x + 2sin X)/(x + 2cos X) = 0
In the fascinating realm of calculus, limits serve as the bedrock upon which the edifice of continuity, derivatives, and integrals is constructed. Understanding limits is not merely an academic exercise; it's a crucial step towards grasping the behavior of functions as their inputs approach specific values. This article embarks on a detailed exploration of two fundamental limits: (i) lim (x→0) (e^x - 1)/x = 1 and (ii) lim (x→0) (x + 2sin x)/(x + 2cos x) = 0. We will dissect each limit, providing rigorous proofs and insightful explanations to illuminate the underlying concepts. By the end of this journey, you will not only comprehend the mechanics of these limits but also appreciate their significance in the broader context of mathematical analysis.
Delving into the Heart of the Limit
At first glance, the limit lim (x→0) (e^x - 1)/x may appear enigmatic. We are tasked with determining the behavior of the function (e^x - 1)/x as x approaches 0. A naive substitution of x = 0 leads to the indeterminate form 0/0, a classic signal that further investigation is required. To unravel this limit, we will employ a multi-pronged approach, leveraging the power of L'Hôpital's Rule, the Maclaurin series expansion of e^x, and a geometric interpretation.
Method 1: L'Hôpital's Rule – A Powerful Tool
L'Hôpital's Rule is a cornerstone of limit evaluation, particularly effective when dealing with indeterminate forms. It states that if the limit of f(x)/g(x) as x approaches c yields 0/0 or ∞/∞, and if the limit of f'(x)/g'(x) exists, then:
lim (x→c) f(x)/g(x) = lim (x→c) f'(x)/g'(x)
In our case, f(x) = e^x - 1 and g(x) = x. Both functions approach 0 as x approaches 0, satisfying the prerequisite for L'Hôpital's Rule. Now, let's compute the derivatives:
f'(x) = d/dx (e^x - 1) = e^x g'(x) = d/dx (x) = 1
Applying L'Hôpital's Rule, we get:
lim (x→0) (e^x - 1)/x = lim (x→0) e^x / 1 = e^0 = 1
Thus, we have successfully demonstrated that lim (x→0) (e^x - 1)/x = 1 using L'Hôpital's Rule. This method showcases the rule's efficacy in transforming an indeterminate form into a readily evaluable limit.
Method 2: Maclaurin Series Expansion – An Elegant Approach
Another illuminating approach to this limit involves the Maclaurin series expansion of the exponential function e^x. The Maclaurin series is a Taylor series expansion of a function about 0, providing a polynomial representation that converges to the function within a certain interval. The Maclaurin series for e^x is:
e^x = 1 + x + x^2/2! + x^3/3! + x^4/4! + ...
Substituting this expansion into our limit expression, we have:
lim (x→0) (e^x - 1)/x = lim (x→0) (1 + x + x^2/2! + x^3/3! + ... - 1) / x
Simplifying, we get:
lim (x→0) (x + x^2/2! + x^3/3! + ...) / x
Now, we can factor out an x from the numerator:
lim (x→0) x(1 + x/2! + x^2/3! + ...) / x
Canceling the x terms (since x is approaching 0 but not equal to 0), we obtain:
lim (x→0) (1 + x/2! + x^2/3! + ...)
As x approaches 0, all terms containing x vanish, leaving us with:
lim (x→0) (1 + x/2! + x^2/3! + ...) = 1
This elegant method, leveraging the Maclaurin series expansion, reinforces our earlier conclusion that lim (x→0) (e^x - 1)/x = 1. It also provides a deeper understanding of the behavior of e^x near 0.
Method 3: Geometric Interpretation – A Visual Insight
To further solidify our understanding, let's explore a geometric interpretation of this limit. Consider the graph of the function e^x. The derivative of e^x at x = 0 represents the slope of the tangent line to the curve at that point. We know that the derivative of e^x is itself e^x, so the slope of the tangent line at x = 0 is e^0 = 1.
Now, let's visualize the expression (e^x - 1)/x. This expression represents the slope of the secant line connecting the points (0, e^0) = (0, 1) and (x, e^x) on the graph of e^x. As x approaches 0, this secant line gets closer and closer to the tangent line at x = 0.
Therefore, the limit of (e^x - 1)/x as x approaches 0 is the slope of the tangent line to e^x at x = 0, which we know is 1. This geometric perspective provides a visual confirmation of our result.
Conclusion for Limit (i)
Through three distinct methods – L'Hôpital's Rule, Maclaurin series expansion, and geometric interpretation – we have unequivocally proven that lim (x→0) (e^x - 1)/x = 1. Each method offers a unique perspective, enriching our comprehension of this fundamental limit. This limit serves as a cornerstone in calculus, appearing in various contexts, including the definition of the derivative of e^x and the derivation of other important limits.
Unraveling the Trigonometric Limit
Our next endeavor is to evaluate the limit lim (x→0) (x + 2sin x)/(x + 2cos x). Unlike the previous limit, direct substitution of x = 0 does not lead to an indeterminate form. However, it's still crucial to analyze the behavior of the function as x approaches 0 to ensure the limit exists and to determine its value.
Direct Substitution – A Straightforward Approach
Let's begin by attempting a direct substitution of x = 0 into the expression (x + 2sin x)/(x + 2cos x):
(0 + 2sin 0) / (0 + 2cos 0) = (0 + 2 * 0) / (0 + 2 * 1) = 0 / 2 = 0
In this case, direct substitution yields a finite value, 0. This strongly suggests that the limit exists and is equal to 0. However, to provide a more rigorous proof, let's delve into an alternative method.
Alternative method: Limit Laws and Squeeze Theorem
We will break down this limit into smaller parts using limit laws. We know the following standard limits:
- lim (x→0) sin(x) = 0
- lim (x→0) cos(x) = 1
We are given:
lim (x→0) (x + 2sin x)/(x + 2cos x)
As x approaches 0:
lim (x→0) (x + 2sin x)/(x + 2cos x) = (0 + 2 * 0)/(0 + 2 * 1) = 0/2 = 0
So, the limit is 0. This direct approach confirms that as x approaches 0, the function (x + 2sin x)/(x + 2cos x) approaches 0.
Conclusion for Limit (ii)
Through a direct substitution and breaking the limit into smaller parts, we have rigorously established that lim (x→0) (x + 2sin x)/(x + 2cos x) = 0. This result underscores the importance of direct substitution as an initial step in limit evaluation. It also highlights how limit laws can simplify complex expressions, making it easier to determine their limiting behavior. This limit, while seemingly simple, exemplifies the interplay between trigonometric functions and algebraic expressions in the realm of calculus.
In this comprehensive exploration, we have rigorously proven two fundamental limits: (i) lim (x→0) (e^x - 1)/x = 1 and (ii) lim (x→0) (x + 2sin x)/(x + 2cos x) = 0. For the first limit, we employed L'Hôpital's Rule, Maclaurin series expansion, and a geometric interpretation, showcasing the multifaceted nature of limit evaluation. For the second limit, a direct substitution provided a straightforward solution, highlighting the importance of this initial step. These limits serve as cornerstones in calculus, underpinning concepts such as derivatives, continuity, and more advanced limit calculations. A thorough understanding of these limits equips us with the tools to navigate the intricacies of mathematical analysis and appreciate the beauty of functional behavior as inputs approach specific values. As we venture further into the world of calculus, these foundational limits will serve as guiding lights, illuminating the path to more complex and profound mathematical concepts.
