Problem 67 - University Calculus: Early Transcendentals

Question

Evaluate each limit by first converting each to a derivative at a particular \(x\) -value. $$\lim _{x \rightarrow 1} \frac{x^{50}-1}{x-1}$$

Step-by-Step Solution

Answer

The limit is 50.

Step 1: Recognize the Derivative Form

The given limit \( \lim_{x \rightarrow 1} \frac{x^{50} - 1}{x - 1} \) resembles the standard form of the definition of a derivative \( \lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a} \) for a particular function \( f(x) \) evaluated at \( x = a \).

Step 2: Identify the Function and Point

Here, the expression can be seen as the derivative of the function \( f(x) = x^{50} \) at the point \( a = 1 \). This is because in \( \frac{x^{50} - 1}{x - 1} \), the form \( f(x) - f(a) \) corresponds to \( x^{50} - 1 \) and \( x - 1 \) represents \( x - a \).

Step 3: Discuss the Derivative

The derivative \( f'(x) \) of the function \( f(x) = x^{50} \) can be found using the power rule, which states that \( \frac{d}{dx}[x^n] = n \cdot x^{n-1} \).

Step 4: Calculate the Derivative

Applying the power rule to \( f(x) = x^{50} \), we have:\[ f'(x) = 50x^{49} \]

Step 5: Evaluate at the Point

Now, evaluate the derivative at the point \( x = 1 \):\[ f'(1) = 50 \times 1^{49} = 50 \]

Step 6: Conclude the Limit

Thus, the original limit is equal to the derivative of \( f \) at \( x = 1 \), which is 50.

Key Concepts

Derivative

A derivative is a fundamental concept in calculus that measures how a function changes as its input changes. It's like finding the slope of a tangent line to the function at a given point. In simpler terms, it tells you the rate of change of the function relative to its variable. For example, if you were to drive a car, the derivative would give you the speedometer reading of your car at any point in time, reflecting how fast you're going. Understanding derivatives helps in predicting the behavior of functions, whether they are increasing, decreasing, or have reached a maximum or minimum. In the given exercise, the problem involves finding the derivative of the function at a particular point by using the limit definition of a derivative.

Power Rule

The power rule is a quick way to find the derivative of a function that is a power of a variable. It's incredibly useful when dealing with polynomials. The rule states that for a function of the form \( f(x) = x^n \), the derivative \( f'(x) \) is \( n \times x^{n-1} \). This makes it easy to calculate how functions behave without repeatedly using the limit definition of a derivative.In our exercise, we applied the power rule to \( f(x) = x^{50} \). Using the rule, the derivative \( f'(x) = 50x^{49} \) was determined. This step simplifies the process, making calculus problems quicker and often easier to solve.

Function Evaluation

Function evaluation is the process of finding the value of a function at a specific input. In the context of derivatives, once we find the derivative \( f'(x) \), we often need to evaluate it at a particular point to understand the function's behavior at that point.For our calculus problem, after finding \( f'(x) = 50x^{49} \), we evaluated the derivative at \( x = 1 \) to solve the limit problem. The function evaluation in step 5 showed that \( f'(1) = 50 \), confirming the value of the limit. This step bridges the calculation of derivatives with practical insights into the function's rate of change at specific points.

Calculus

Calculus is a branch of mathematics that studies how things change. It has two main branches: differentiation and integration. The derivative, which we calculated using the power rule in our exercise, is a part of differentiation. Calculus allows us to solve problems involving rates of change and areas under curves, among others. It’s vital in various fields such as physics, engineering, economics, and biology. By mastering calculus concepts like derivatives, you gain powerful tools to analyze the real world and solve complex problems. The given problem used calculus principles to determine how a high-powered function behaves near a particular point. Understanding these concepts makes solving real-world problems both effective and satisfying.