Problem 37 - Precalculus

Question

Evaluate the following limits. Write your answer in simplest form. $$\lim _{h \rightarrow 0} \frac{\left[2(x+h)^{2}-(x+h)\right]-\left(2 x^{2}-x\right)}{h}$$

Step-by-Step Solution

Answer

The limit is \(4x - 1\).

Step 1: Expand the Expression

The given expression is \[2(x+h)^2 - (x+h) - (2x^2-x)\]. First, expand \((x+h)^2\) as \(x^2 + 2xh + h^2\). Then, plug this into the expression to get \[2(x^2 + 2xh + h^2) - (x+h) - (2x^2-x)\].

Step 2: Simplify the Expression

Expand and simplify within each bracket. Calculate \[2(x^2 + 2xh + h^2) = 2x^2 + 4xh + 2h^2\]. Combine like terms across the entire expression to yield: \[2x^2 + 4xh + 2h^2 - x - h - 2x^2 + x\].

Step 3: Combine Like Terms

Combine all like terms: cancel out \(2x^2\) and \(-2x^2\) as well as \(-x\) and \(x\). What's left is \[4xh + 2h^2 - h\].

Step 4: Factor and Simplify

Factor out \(h\) from the expression: \[h(4x + 2h - 1)\]. This makes the expression \(\frac{h(4x + 2h - 1)}{h}\).

Step 5: Apply the Limit

Since \(h\) cancels out, apply the limit as \(h \to 0\), resulting in \[4x + 2(0) - 1 = 4x - 1\].

Key Concepts

Difference Quotient

The difference quotient is at the heart of calculus, allowing us to understand how functions behave as they approach specific points. In the given exercise, the difference quotient is expressed as:

• \(\frac{f(x+h) - f(x)}{h}\)

This formula efficiently captures the idea of the rate of change of the function \(f\) as \(h\), a small increment, approaches zero. Here, \(f(x)\) is a polynomial function, making it crucial to expand and simplify accurately to proceed with limit evaluation.

Limit Evaluation

Limit evaluation is essential in understanding the behavior of functions when a certain variable approaches a particular value. In our case, we are finding the limit as \(h\) approaches zero for the expression:

• \(\lim_{h \to 0} \frac{[2(x+h)^2 - (x+h)] - (2x^2 - x)}{h}\)

During this process, it's important first to simplify the expression so that any initial indeterminate forms like \(\frac{0}{0}\) are resolved. By simplifying, you can directly evaluate the limit without causing algebraic complications.

Simplifying Expressions

Simplifying expressions is a fundamental skill in algebra and calculus. This involves expanding squared terms, combining like terms, and factoring.

• First, expand \((x+h)^2\) to obtain \(x^2 + 2xh + h^2\).
• Substitute back into the initial expression to expand it further.
• After expansion, simplify by cancelling out terms like \(2x^2\) to reduce complexity.

This makes subsequent steps, like factoring, more straightforward and helps in the eventual evaluation of the limit.

Binomial Expansion

Binomial expansion allows us to break down expressions like \((x+h)^2\) into simpler terms:

• \(x^2 + 2xh + h^2\)

Using binomial expansion in calculus exercises helps detail the changes occurring in polynomial expressions. Correctly applying the expansion ensures no critical parts of the expression are omitted or incorrectly adjusted, allowing for a clearer simplification path when dealing with limits.

Limit Application

Applying the limit in calculus is the final step after simplification. Here, as \(h\) approaches zero, ensure any remaining terms containing \(h\) are eliminated so the expression simplifies correctly.

• In the expression \(h(4x + 2h - 1)\), \(h\) cancels with the denominator.
• Plug \(h = 0\) into the simplified expression: \(4x + 2(0) - 1\).
• This gives the final solution: \(4x - 1\).

By successfully reducing the difference quotient to this expression, the limit shows the rate of change of the original function, reinforcing our understanding of derivatives.