Question

find the limit $$ \lim _{\Delta x \rightarrow 0} \frac{2(x+\Delta x)-2 x}{\Delta x} $$

Short answer

The value of the limit is 2.

Step-by-step solution

Simplify the Numerator

Attempt to simplify the expression by simplifying its numerator. Begin by simplifying \(2(x+\Delta x)-2 x\). Multiply out the brackets to give \(2x+2\Delta x-2x\). This simplifies to \(2\Delta x\). Our limit therefore simplifies to: \( \lim _{\Delta x \rightarrow 0} \frac{2\Delta x}{\Delta x} \)

Cancel Common Terms

Note that \( \Delta x \) appears in both the numerator and denominator. Cancel \( \Delta x \) from both the numerator and the denominator. Our limit now becomes \( \lim _{\Delta x \rightarrow 0} 2 \)

Compute the Limit

Now that the indeterminate form has been removed, the limit can be computed directly. However, this limit is simply 2 since 2 is a constant and the limit of a constant is simply the constant itself. Therefore, \( \lim _{\Delta x \rightarrow 0} 2 = 2 \)

Key concepts

Simplifying Expressions

When faced with finding the limit of a function as a variable approaches a certain value, simplifying the expression is crucial to make the problem more manageable.
Simplification includes expanding brackets, combining like terms, and breaking down complex fractions. In our exercise, the process began with expanding the expression within the brackets, resulting in terms that included both the initial variable (\(x\)) and the increment or change (\( \(Delta x \) \) ). By combining like terms, we're often left with a much less cluttered, more straightforward expression, facilitating the application of limit laws and rules.

Canceling Common Terms

Upon simplification, we often encounter expressions where the same term appears in both the numerator and the denominator. These are 'common terms,' which can be canceled out for further simplification.
In the context of limits, canceling out common terms is critical, especially when these terms contain the variable that approaches a certain value. The presence of this variable often signifies an indeterminate form, which canceling can resolve. In our exercise, '\( \(Delta x \) \) ' was canceled from both the numerator and the denominator, streamlining the expression and bringing us a step closer to finding the limit.

Indeterminate Forms

Indeterminate forms are expressions in calculus that do not directly reveal the limit and can take on different values depending on variables' behavior. Common examples include 0/0 and ∞/∞.
In this exercise, before simplification and canceling, it appears that as '\( \(Delta x \) \) ' approaches zero, our expression would have been undefined, hinting at an indeterminate form. However, simplification and canceling helped us avoid this pitfall by removing the variable, leaving us with a constant. Since the limit of a constant is just the constant itself, resolving the indeterminate form showed that the limit is indeed 2.