Problem 29 - Contemporary Precalculus

Question

Find the limit if it exists. If the limit does not exist, explain why. \(\lim _{x \rightarrow 0}\left(\frac{1 /(x+5)-1 / 5}{x}\right) \quad[\text {Hint}:\) Write the expression in parentheses as a single fraction.]

Step-by-Step Solution

Answer

\) Answer: The limit is \(\frac{-1}{25}\).

Step 1: Simplify the expression inside the limit

To simplify the given expression, we can first find a common denominator for the fractions \(1/(x+5)\) and \(1/5\). Then, subtract these fractions and divide by x. The common denominator for these fractions is \((x+5)\cdot5\). Thus, we rewrite the expression inside the parentheses as: \(\frac{1/(x+5)-1/5}{x}=\frac{5- (x+5)}{5x(x+5)}\) Next, we simplify the numerator: \(5-(x+5)=5-x-5=-x\) So, our simplified expression inside the limit becomes: \(\lim_{x\rightarrow 0}\left(\frac{-x}{5x(x+5)}\right)\)

Step 2: Cancel out the x term

We can further simplify the expression by canceling out the x term from both the numerator and the denominator: \(\lim_{x\rightarrow 0}\left(\frac{-x}{5x(x+5)}\right)=\lim_{x\rightarrow 0}\left(\frac{-1}{5(x+5)}\right)\)

Step 3: Apply the limit

Now that our expression is simplified, we can apply the limit as x approaches 0: \(\lim_{x\rightarrow 0}\left(\frac{-1}{5(x+5)}\right)= \frac{-1}{5(0+5)}\)

Step 4: Solve for the limit

Finally, we solve for the limit: \(\lim_{x\rightarrow 0}\left(\frac{-1}{5(x+5)}\right)= \frac{-1}{5(0+5)}=\frac{-1}{5\cdot5}=\frac{-1}{25}\) The limit exists and is equal to \(\frac{-1}{25}\).

Key Concepts

Limits

Understanding limits is a fundamental concept in calculus. Limits help us study the behavior of functions as the input approaches a certain value. When tackling limits, it's essential to focus on what happens as x gets arbitrarily close to a particular point, called the limit point. In our given exercise, we're examining what happens as x approaches 0.
The key idea is to see how the function behaves near this point without necessarily having to reach it. This way, we can determine if there's a specific value that the function gets infinitely close to as the input becomes very small or very large.
In practical terms, applying limits often involves creating equivalent and simpler forms of expressions. This ensures that any indeterminate forms, such as those arising from 0/0 calculations, can be resolved.

Simplifying Expressions

Simplifying expressions is a crucial skill, especially when dealing with limits. Our goal is to rewrite complex expressions in a simpler form that is easier to analyze. In the original problem, we started with \[\lim _{x \rightarrow 0}\left(\frac{1 /(x+5)-1 / 5}{x}\right)\]and needed to turn the expression inside the parentheses into a single fraction:

• Found a common denominator: \((x+5)\cdot5\).
• Rewrote the expression: \[\frac{5 -(x+5)}{5x(x+5)}\].
• Simplified the numerator: \(-x\).

This process turned a complex difference of two fractions into a much simpler single fraction, \[\frac{-x}{5x(x+5)}\],which could be further simplified by cancelling terms, leading us closer to calculating the limit.

Fractions

Working with fractions often involves combining separate fractions into a single entity. This can be challenging when fractions have different denominators. In solving for limits, having a single fraction simplifies the analysis greatly. Here are some helpful strategies:

• Identify the least common denominator to combine separate fractions.
• Rewrite each fraction so they share this common denominator.
• Perform operations like subtraction or addition now that the denominators match.

In our example, we needed to handle the difference \(\frac{1}{x+5}\) and \(\frac{1}{5}\).The common denominator here was \((x+5)\times5\).
After simplification, fractions can often be reduced by cancellation, applying both algebraic understanding and simplifying rules to get closer to an analyzable form.