Question

L'Hôpital's Rule in its 1696 form said this: If \(\lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a} g(x)=0\), then \(\lim _{x \rightarrow a} f(x) / g(x)=f^{\prime}(a) / g^{\prime}(a)\) provided that \(f^{\prime}(a)\) and \(g^{\prime}(a)\) both exist and \(g^{\prime}(a) \neq 0\). Prove this result without recourse to Cauchy's Mean Value Theorem.

Short answer

L'Hôpital's Rule is proven by expressing the limit as a ratio of derivatives using the definition of derivatives.

Step-by-step solution

Define the Problem

We need to prove L'Hôpital's Rule which states that \(\lim _{x \rightarrow a} \frac{f(x)}{g(x)} = \frac{f^{\prime}(a)}{g^{\prime}(a)}\) under the conditions \(\lim_{x \rightarrow a} f(x) = 0\) and \(\lim_{x \rightarrow a} g(x) = 0\), and both derivatives \(f^{\prime}(a)\) and \(g^{\prime}(a)\) exist with \(g^{\prime}(a) eq 0\).

Consider the Basic Limit Form

Since \(f(x)\) and \(g(x)\) approach zero as \(x\) approaches \(a\), we have an indeterminate form \(\frac{0}{0}\). L'Hôpital's Rule suggests a way to evaluate this by differentiating the numerator and the denominator.

Use the Limit Definition of Derivatives

Recall that the derivative of a function at a point is defined as \(f^{\prime}(a) = \lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a}\). Since \(f(a) = 0\) (as given by the problem's conditions), this simplifies to \(\lim_{x \rightarrow a} \frac{f(x)}{x - a}\). Similarly, for \(g(x)\), \(g^{\prime}(a) = \lim_{x \rightarrow a} \frac{g(x)}{x - a}\) since \(g(a) = 0\).

Apply the Limits to the Ratio

Consider the ratio \(\frac{f(x)}{g(x)}\). We can express this as the product of two limits: \[\frac{f(x)}{g(x)} = \frac{f(x)}{x-a} \cdot \frac{x-a}{g(x)}\]. The limit of this product as \(x\) approaches \(a\) is the product of the individual limits, \[\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f(x)}{x-a} \times \lim_{x \rightarrow a} \frac{x-a}{g(x)}\].

Substitute the Derivatives

Using the derivative limits defined previously, substitute \(\lim_{x \rightarrow a} \frac{f(x)}{x-a} = f^{\prime}(a)\) and \(\lim_{x \rightarrow a} \frac{x-a}{g(x)} = \frac{1}{g^{\prime}(a)}\). So, \[\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = f^{\prime}(a) \times \frac{1}{g^{\prime}(a)} = \frac{f^{\prime}(a)}{g^{\prime}(a)}\].

Conclude the Proof

We have shown that under the given conditions, the limit of the ratio of the functions equals the ratio of their derivatives. This concludes the proof of L'Hôpital’s Rule without using Cauchy's Mean Value Theorem.

Key concepts

Limit Definition of Derivatives

Understanding the limit definition of derivatives is essential in calculus. The derivative of a function at a point measures how the function's value changes as its input changes slightly. Mathematically, if we have a function \(f(x)\), its derivative at a particular point \(a\) is defined as:\[f^{\prime}(a) = \lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a}\]This formula represents the slope of the tangent line to the curve of the function at point \(a\). It gives us the rate at which \(f(x)\) changes. In cases where the function equals zero at a point (\(f(a) = 0\)), this simplifies to:\[f^{\prime}(a) = \lim_{x \rightarrow a} \frac{f(x)}{x - a}\]This simplification helps when dealing with limits in the context of L'Hôpital's Rule, as we can apply the derivative limits directly.

Indeterminate Form 0/0

Indeterminate forms such as \(0/0\) arise frequently in calculus, especially in limit problems. An indeterminate form is a fraction where both the numerator and the denominator tend toward zero, creating uncertainty about the overall limit.In simpler terms, if both \(f(x)\) and \(g(x)\) approach zero as \(x\) approaches \(a\), we face a situation of \(\frac{0}{0}\). Without a tool like L'Hôpital's Rule, it's challenging to evaluate such limits directly. This is because multiplying and dividing by zero is indeterminate and does not give any useful information.L'Hôpital's Rule comes into play by allowing us to differentiate the numerator and the denominator separately, turning an indeterminate form into something solvable with standard limit techniques.

Proof in Calculus

Proving calculus concepts can initially seem daunting, but breaking down the steps can illuminate complex theories. A proof provides logical reasoning that certain mathematical assertions hold true under given conditions.For instance, in proving L'Hôpital's Rule without Cauchy's Mean Value Theorem, we follow a structured approach:

• Start with the problem definition and known conditions, such as \(\lim_{x \rightarrow a} f(x) = 0\) and \(\lim_{x \rightarrow a} g(x) = 0\).


• Recognize the indeterminate form and apply the limit definition of derivatives to each function.


• Utilize the limits' properties to switch the problem from \(f(x)/g(x)\) to derivative forms \(f^{\prime}(a)/g^{\prime}(a)\).


• Verify that these steps hold under the constraints of the problem, such as ensuring \(g^{\prime}(a) eq 0\).

This structured reasoning and verification form the core of any robust proof in calculus.

Derivatives of Functions

Derivatives, which represent the rate of change, are crucial across all of calculus. They serve as the backbone for concepts like L'Hôpital's Rule.When differentiating functions, say \(f(x)\) and \(g(x)\), you obtain their respective derivatives \(f^{\prime}(x)\) and \(g^{\prime}(x)\). These derivatives help us evaluate limits that are initially expressed in indeterminate forms. In essence, we transform a complex limit into a derived ratio that's easier to understand and solve.The application of derivatives in solving real-world problems cannot be overstated. Whether it's understanding motion, predicting economic trends, or designing curves in graphics, derivatives offer insights into how things vary.Remember:

• Derivatives convert unknown limit forms into finite values.


• They indicate whether a function increases or decreases at any given point.


• Regular use of differentiation strengthens problem-solving abilities in mathematics.