Problem 60 - Single Variable Calculus

Question

Find the limit, if it exists. \(\lim _{h \rightarrow 0} \frac{(2+h)^{-1}-2^{-1}}{h}\)

Step-by-Step Solution

Answer

The short answer to the given limit is \(\lim_{h \rightarrow 0} \frac{(2+h)^{-1}-2^{-1}}{h} = -\frac{1}{4}\).

Step 1: Rewrite the expression with common denominators

The expression is given as \(\frac{(2+h)^{-1}-2^{-1}}{h}\). To simplify this expression, we can rewrite it by finding a common denominator for both \( (2 + h)^{-1} \) and \( 2^{-1} \). \(\frac{(2+h)^{-1}-2^{-1}}{h} = \frac{\frac{1}{2+h} - \frac{1}{2}}{h}\)

Step 2: Make the denominators the same and simplify the fraction

Now, we can make the denominators the same by multiplying both numerator and denominator of the fractions by the opposite denominator: \(\frac{\frac{1}{2+h} - \frac{1}{2}}{h} = \frac{\frac{2 - (2+h)}{2(2+h)}}{h}\) Now, simplify the numerator: \(\frac{\frac{-h}{2(2+h)}}{h}\)

Step 3: Cancel out the h terms

Now, cancel out the h in the numerator and denominator: \(\frac{\frac{-h}{2(2+h)}}{h} = \frac{-1}{2(2+h)}\)

Step 4: Evaluate the limit

Now we can take the limit as h approaches 0: \(\lim_{h \rightarrow 0} \frac{-1}{2(2+h)} = \frac{-1}{2(2+0)} = \frac{-1}{4} \) So the limit is \(\boxed{-\frac{1}{4}}\).

Key Concepts

Limit of a Function

The concept of the limit of a function is fundamental in calculus and serves as the cornerstone of many of the subject's principles, including derivatives and integrals. A limit describes the value that a function approaches as the input (or the variable) approaches some value.

For example, in the expression \(\lim _{h \rightarrow 0} \frac{(2+h)^{-1}-2^{-1}}{h}\), we are interested in finding what value the function \(\frac{(2+h)^{-1}-2^{-1}}{h}\) approaches as \(h\) gets infinitely close to zero. The limit does not always equal the function's actual value at that point (if it even exists), but rather the value the function is approaching.

Understanding the behavior of functions as they get close to a certain point allows mathematicians to discuss and compute rates of change, continuous growth, and areas under curves, which are all core concepts of calculus utilized in various fields such as physics, engineering, and economics.

Simplifying Expressions

Simplifying expressions is a valuable skill in calculus, as it often makes the evaluation of limits, derivatives, and integrals much more manageable. The process typically involves reducing the complexity of an expression without changing its value. For the limit in question, we simplify \(\frac{(2+h)^{-1}-2^{-1}}{h}\) by finding a common denominator.

When simplifying, it's essential to:

• Combine like terms.
• Factor expressions when possible.
• Cancel out terms where appropriate.

Our simplification makes the subsequent cancellation possible, leading to a straightforward evaluation of the limit. Simplifying not only makes calculations easier but also aids in recognizing patterns and understanding the behavior of functions.

Common Denominators

Finding common denominators is a technique used to combine fractions that have different denominators. This is crucial for addition or subtraction of fractions, and in the context of calculus, for simplifying complex rational expressions before taking a limit.

To apply this method, each fraction's numerator and denominator is often multiplied by the denominator of the other fraction, creating equivalent fractions with a shared common denominator. In the given exercise, the common denominator for \(\frac{1}{2+h}\) and \(\frac{1}{2}\) is \(2(2+h)\), which allows us to combine the fractions within the limit expression into a single fraction. Once we have a common denominator, it often reveals a simplification that leads to the elimination of terms, especially variables in the denominator that might cause the expression to be undefined at certain points, as seen with the cancelling of \(h\) terms in our case.