Question

Suppose \(f^{\prime}(x), g^{\prime}(x)\) exist, \(g^{\prime}(x) \neq 0\), and \(f(x)=g(x)=0\). Prove that $$ \lim _{t \rightarrow x} \frac{f(t)}{g(t)}=\frac{f^{\prime}(x)}{g^{\prime}(x)} $$ (This holds also for complex functions.)

Short answer

Question: Prove that if the derivative of both functions f(x) and g(x) exist, g'(x) ≠ 0, and both functions equal zero at x, then the limit of the ratio of the two functions as we approach x is equal to the ratio of their derivatives at that point. Answer: By L'Hôpital's Rule, we can prove that the limit of the ratio of the two functions equals the ratio of their derivatives when both functions equal zero at a point x: $$ \lim_{t \to x} \frac{f(t)}{g(t)} = \frac{f'(x)}{g'(x)} $$

Step-by-step solution

Use L'Hôpital's Rule

Given that the limit of both functions \(f(t)\) and \(g(t)\) goes to 0 as \(t\) approaches \(x\), we can apply L'Hôpital's Rule: $$ \lim _{t \rightarrow x} \frac{f(t)}{g(t)}=\lim_{t \rightarrow x} \frac{f^{\prime}(t)}{g^{\prime}(t)} $$

Apply the definition of the derivative

We can re-write the derivatives \(f'(t)\) and \(g'(t)\) using the definition of the derivative: $$ f^{\prime}(t)=\lim _{h \rightarrow 0} \frac{f(t+h)-f(t)}{h} $$ $$ g^{\prime}(t)=\lim _{h \rightarrow 0} \frac{g(t+h)-g(t)}{h} $$ Substitute the definitions of the derivatives back into the limit of the ratio: $$ \lim _{t \rightarrow x} \frac{f^{\prime}(t)}{g^{\prime}(t)} =\lim _{t \rightarrow x} \frac{\lim_{h\to0}\frac{f(t+h)-f(t)}{h}}{\lim_{h\to0}\frac{g(t+h)-g(t)}{h}} $$

Combine the limits under one limit statement

Since we are taking the limit as \(t \rightarrow x\) and then the limit as \(h \rightarrow 0\), we can combine the two limits under a single limit statement, as long as both limits exist: $$ \lim _{t \rightarrow x} \frac{f^{\prime}(t)}{g^{\prime}(t)} = \lim _{t \rightarrow x, h \rightarrow 0} \frac{\frac{f(t+h)-f(t)}{h}}{\frac{g(t+h)-g(t)}{h}} $$

Simplify the expression

Now we can simply cancel out the \(h\)'s in the numerator and the denominator to simplify the expression: $$ \lim _{t \rightarrow x, h \rightarrow 0} \frac{f(t+h)-f(t)}{g(t+h)-g(t)} $$

Analyze the limit as \(t \rightarrow x\) and the expression simplifies

As \(t\) approaches \(x\), the expression simplifies to: $$ \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{g(x+h)-g(x)} $$ Since we know from the problem statement that \(f(x)=g(x)=0\), the expression becomes: $$ \lim _{h \rightarrow 0} \frac{f(x+h)}{g(x+h)} $$

Analyze the limit as \(h \rightarrow 0\)

Finally, as \(h\) approaches 0, both \(x+h\) and \(x\) will have the same values and the expression will become: $$ \frac{f^{\prime}(x)}{g^{\prime}(x)} $$ Since we have simplified the original limit expression and obtained the same ratio of derivatives, we have proven the statement: $$ \lim _{t \rightarrow x} \frac{f(t)}{g(t)}=\frac{f^{\prime}(x)}{g^{\prime}(x)} $$

Key concepts

Limits of Functions

Understanding the concept of a limit is fundamental in calculus, and it plays a critical role when dealing with functions whose values approach a certain point. A limit captures the behavior of a function as the input approaches a particular value, without necessarily reaching that value. For instance, when we say
\( \lim_{x \to c} f(x) = L \),
we're describing the tendency of \( f(x) \) to get closer and closer to \( L \) as \( x \) approaches \( c \). It's not about where the function is at \( x=c \), but where it's headed as \( x \) gets indefinitely close to \( c \).
The exercise provided involves limits where both the numerator and the denominator of a fraction go to zero, which is known as an indeterminate form. L'Hôpital's Rule is often used to resolve this by taking derivatives, provided certain conditions are met. It allows us to convert a difficult-to-evaluate limit into one that is easier to handle by examining the rates of change (derivatives) of the top and bottom of the fraction separately.

Derivatives

In calculus, the derivative represents the rate at which a function changes at any point. Mathematically, it's defined as the limit of the average rate of change of the function over a tiny interval as that interval approaches zero. The formal definition for the derivative of a function \( f \) at a point \( x \) is:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
This definition was used in the step-by-step solution to rewrite the derivatives of functions \( f(t) \) and \( g(t) \). Derivatives are the building blocks for more complex applications in mathematical analysis, enabling the calculation of slopes of tangent lines, rates of change in physics, and optimization problems in engineering and economics. They are especially important when dealing with the indeterminate forms of limits, as they allow application of L'Hôpital's Rule to simplify otherwise challenging limit calculations.

Mathematical Analysis

Mathematical analysis is a broad area of mathematics that includes the study of limits, continuity, derivatives, and integrals. It's essentially an extension of calculus that delves into proofs and theorems to provide a solid foundation to these concepts, ensuring they are rigorously defined and understood. Analysis not only deals with real numbers but also covers complex functions and higher dimensions.
The exercise involves an aspect of analysis where we prove a limit statement using precise definitions. We're not just computing a limit; we're justifying it according to the principles of analysis. Throughout the steps of the solution, the limit is manipulated algebraically and logically to verify the initial condition. The exercise showcases analysis in action, combining the concepts of limits and derivatives to establish a deeper understanding of how functions behave near a point of interest and how they can be compared through their rates of change.