Problem 59 - Calculus of a Single Variable

Question

Finding a Limit In Exercises \(47-62,\) find the limit. $$ \lim _{\Delta x \rightarrow 0} \frac{2(x+\Delta x)-2 x}{\Delta x} $$

Step-by-Step Solution

Answer

The limit of the function as \(\Delta x\) approaches 0 is 2.

Step 1: Simplify the function

First, simplify the fraction in the limit. The function \(\frac{2(x+\Delta x)-2x}{\Delta x}\) simplifies to \(\frac{2x+2\Delta x-2x}{\Delta x}\), which simplifies further to \(\frac{2\Delta x}{\Delta x}\). Here, the 2\(\Delta x\) terms in the numerator are identical and cancel each other out, leaving you with a function of \(2\).

Step 2: Substitute the limit

After we simplify the function we are left with a constant, \(2\). The limit of a constant is always the constant itself. In other words, substituting \(\Delta x\) for 0 won't affect the results because there are no more \(\Delta x\) terms in the function.

Step 3: Calculate the limit

Substituting \(\Delta x\) for 0 in our function of \(2\), the function remains \(2\). So, \(\lim _{\Delta x \rightarrow 0} \frac{2(x+\Delta x)-2 x}{\Delta x} = 2\).

Key Concepts

Simplifying Expressions

Simplifying expressions is a fundamental step in solving many calculus problems, particularly when dealing with limits. When we look at the original problem, we have \( \lim _{\Delta x \rightarrow 0} \frac{2(x+\Delta x)-2 x}{\Delta x} \). Simplification involves breaking down this complex expression into a simpler form. By expanding the numerator, we see it becomes \(2x + 2\Delta x - 2x\), resulting in \(2\Delta x\). Simplifying further, the \(2x\) terms cancel out, leaving us with \(\frac{2\Delta x}{\Delta x}\). This step is crucial because it allows us to see that the entire expression simplifies to a constant \(2\), eliminating \(\Delta x\) from the denominator and numerator, which is essential for the next step of evaluating the limit.

Substitution Method

The substitution method is often used to find limits, especially after simplification has reduced the function to a manageable form. After simplifying the expression to \(2\), there are no \(\Delta x\) terms left. Thus, substituting \(\Delta x = 0\) directly would have no effect on the value of the expression. In using the substitution method for limits, if the function simplifies to a constant, the limit of the function is simply the same constant. In this particular exercise, even if \(\Delta x\) approaches zero, it doesn’t affect the constant value of \(2\) that the function has simplified to. This makes the substitution method straightforward as the final limit, \(\lim _{\Delta x \rightarrow 0} \frac{2(x+\Delta x)-2 x}{\Delta x} = 2\), is directly derived from the simplified form.

Calculus of a Single Variable

Calculus of a single variable focuses on functions and their limits, derivatives, and integrals, inspecting how these elements behave as input values approach specific points. In this exercise, the focus is on finding the limit by breaking down and simplifying the expression of a function in one variable, \(x\). The single variable calculus aims to study functions just like \( \frac{2(x+\Delta x)-2 x}{\Delta x} \), where \(\Delta x\) represents a small change in \(x\). Here, the goal is to see what happens to the overall expression as this change approaches zero. Calculus of a single variable allows us not only to find this limit but also to understand the behavior and properties of functions near a particular point. This foundational concept is critical in not just limits, but also in understanding the broader picture of differentiation and integration that further explore how functions change and accumulate.