Question

Finding a Limit In Exercises \(7-26\) , find the limit (if it exists). If it does not exist, explain why. $$ \lim _{\Delta x \rightarrow 0^{+}} \frac{(x+\Delta x)^{2}+x+\Delta x-\left(x^{2}+x\right)}{\Delta x} $$

Short answer

\lim _{\Delta x \rightarrow 0^{+}} \frac{(x+\Delta x)^{2}+x+\Deltax-\left(x^{2}+x\right)}{\Delta x} = 2x+1

Step-by-step solution

Simplify the numerator

In the given limit function, the nominator simplifies to \( (x+\Delta x)^2 - x^2 + (x+\Delta x) - x \). Expanding these terms will simplify the numerator to \( x^2+2x\Delta x+(\Delta x)^2 -x^2 + \Delta x + x - x \). This simplification results in \( 2x\Delta x+(\Delta x)^2+ \Delta x \)

Cancel out the common factors

Now that the numerator is simplified, we see that every term has a factor of \( \Delta x \). This common factor can thus be canceled out with the denominator. After canceling out \( \Delta x \), we are left with \( \lim _{\Delta x \rightarrow 0^{+}} (2x+\Delta x+1) \).

Compute the resulting limit

Now, the resulting function is a polynomial function, which is continuous for every real value of x. Therefore, we can directly substitute \( \Delta x = 0 \) into the equation to find the limit. After substitution, we get the limit as \( 2x+1 \)

Key concepts

Limit of a Function

Understanding the limit of a function is a foundational concept in calculus. Essentially, it represents the value that a function approaches as the input (or variable) gets closer to a certain point. In the exercise provided, you are tasked with finding the limit as \( \Delta x \) approaches zero from the positive side, for a particular function.

The process involves simplifying the expression to determine what value the function is moving towards. This specific limit helps in understanding the rate of change, or the derivative, of the function at a point. Calculating the limit can frequently lead to insights about the function’s behavior near the intended point, even if the function itself isn’t well-definied at that exact point.

Continuity

A critical concept linked to limits is continuity. A function is continuous at a point if it meets three criteria: the function is defined at that point, the limit exists at that point, and the limit equals the function’s value at that point.

In our textbook solution, we are working with a polynomial, which is inherently continuous everywhere — meaning, their limits exist at all points and equal the function’s value. Therefore, when finding the limit, you can often substitute the value directly for polynomials. This property significantly simplifies the process of finding limits for continuous functions and provides an intuitive understanding of how a function behaves around a given value.

Polynomials

The given function in the exercise involves a polynomial. Polynomials are algebraic expressions made up of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents. Because of their structure, polynomials are smooth and continuous shapes when graphed.

The simplicity of polynomials not only makes them continuous but also makes simplifying expressions and computing limits straightforward. Since they do not have any jumps or holes, the limit at any point for a polynomial will just be the value of the polynomial at that point, as we see with the function \( 2x+1 \) after simplifying the given expression.

Simplifying Expressions

A crucial skill in working through calculus problems is simplifying expressions. It involves reducing an expression to its most basic form by performing arithmetic operations and combining like terms. Often, simplification can reveal patterns or properties that aren't immediately obvious.

In the context of the given problem, simplifying the numerator involved expanding quadratic terms and combining like terms. This process highlights that every term has a \( \Delta x \) factor, which could then be canceled out with the denominator. Without simplification, identifying that common factor and simplifying the limit would be much more challenging, thus proving the importance of this skill in solving calculus problems.