Problem 30 - Precalculus: Mathematics for Calculus

Question

Finding Limits Evaluate the limit if it exists. $$\lim _{t \rightarrow 0}\left(\frac{1}{t}-\frac{1}{t^{2}+t}\right)$$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Rewrite the Expression

We start by rewriting the given expression for clarity. The expression inside the limit is \( \frac{1}{t} - \frac{1}{t^2 + t} \). To simplify this expression, find a common denominator, which is \(t(t+1)\).

Step 2: Simplify the Expression

The common denominator for the fractions \( \frac{1}{t} \) and \( \frac{1}{t^2 + t} \) is \( t(t+1) \). Thus, we rewrite each fraction: \( \frac{1}{t} = \frac{t+1}{t(t+1)} \) and \( \frac{1}{t^2+t} = \frac{1}{t(t+1)} \). This gives us \( \frac{t+1}{t(t+1)} - \frac{1}{t(t+1)} = \frac{t+1-1}{t(t+1)} = \frac{t}{t(t+1)} \).

Step 3: Cancel and Simplify Further

The expression \( \frac{t}{t(t+1)} \) simplifies by canceling the \( t \) in the numerator and the denominator: \( \frac{1}{t+1} \).

Step 4: Evaluate the Limit

Now substitute \( t = 0 \) into the simplified expression \( \frac{1}{t+1} \). The expression becomes \( \frac{1}{1} = 1 \).

Step 5: Conclusion

Since we have evaluated the limit and found it to be a finite number, the limit exists. Thus, \( \lim_{t \to 0} \left( \frac{1}{t} - \frac{1}{t^2+t} \right) = 1 \).

Key Concepts

Limit Calculation

Calculating limits is a fundamental concept in calculus, allowing us to understand how a function behaves as it approaches a particular point. In the exercise given, the task is to find the limit of the function as the variable \( t \) approaches 0.
This involves simplifying the expression to make direct substitution possible, which is often necessary when the initial form yields an indeterminate form, like \( \frac{0}{0} \). A limit essentially predicts the output value of a function as its input approaches a particular point.

• The limit process involves taking what seems chaotic or undefined and turning it into a precise value.
• This calculation relies on the manipulation of algebraic expressions to remove any "indeterminacies."
• Understanding this gives insight into continuity and the behavior of functions close to the point in question.

By simplifying or rewriting the expression, often the obstacles preventing straightforward substitution are eliminated, enabling limit calculation.

Rational Functions

Rational functions, like the one employed in this limit problem, are quotients of polynomials. In this exercise, the expression \( \frac{1}{t} - \frac{1}{t^2+t} \) is a classic example.
It is crucial to understand how rational functions behave because they often exhibit asymptotic behavior, discontinuities, or holes at specific points, such as \( t = 0 \) in this exercise. This equips us with knowledge about how a function behaves near points where it is not defined.

• Rational expressions often require finding common denominators to combine terms, as shown when rewriting the original expression in the problem.
• Simplifying the expression allows for the identification and cancellation of any factors responsible for indeterminate forms.
• Knowledge of how these functions work can reveal limits' existence and value more effectively.

Understanding rational functions is critical for more advanced calculus problems involving limits, derivatives, and integrals.

Simplifying Expressions

Simplifying expressions is a critical step in solving algebraic problems, especially when dealing with limits. The original function, \( \frac{1}{t} - \frac{1}{t^2+t} \), required finding a common denominator, \( t(t+1) \), to simplify further.
Simplification enables the cancellation of terms that cause indeterminacies like \( \frac{0}{0} \) when limits are evaluated. By canceling terms, we removed the problematic numerator and denominator, making the limit calculable through direct substitution.

• Combining fractions with a common denominator helps consolidate expressions to a simpler form.
• Cancelling terms is a key strategy, particularly when evaluating algebraic expressions to reveal underlying functions.
• Simplifying provides the clarity needed to compute limits accurately and efficiently with fewer computational errors.

Mastering simplification is not only essential for limit problems but also for understanding calculus concepts more broadly.