Problem 22 - Precalculus Mathematics for Calculus

Question

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: Write the Given Limit Expression

The given limit is \( \lim _{t \rightarrow 0}\left(\frac{1}{t}-\frac{1}{t^{2}+t}\right) \). To evaluate this limit, we should first attempt to combine the two fractions into a single expression to simplify our calculations.

Step 2: Finding a Common Denominator

The common denominator for the fractions \( \frac{1}{t} \) and \( \frac{1}{t^{2}+t} \) is \( t(t + 1) \). Use this common denominator to combine the fractions: \[ \frac{1}{t} - \frac{1}{t^{2}+t} = \frac{t+1}{t(t+1)} - \frac{1}{t(t+1)} = \frac{t+1 - 1}{t(t+1)} = \frac{t}{t(t+1)}. \]

Step 3: Simplify the Expression

The expression \( \frac{t}{t(t+1)} \) can be simplified by canceling out \( t \) in the numerator and the denominator: \[ \frac{t}{t(t+1)} = \frac{1}{t+1}. \] This leaves us with a simple expression \( \lim_{t \rightarrow 0} \frac{1}{t+1} \).

Step 4: Evaluate the Limit

Substitute \( t = 0 \) into the simplified expression \( \frac{1}{t+1} \), which gives \( \frac{1}{1} = 1 \). Therefore, \[ \lim_{t \rightarrow 0} \frac{1}{t+1} = 1. \]

Step 5: Conclude the Solution

We have found that the limit \( \lim_{t \rightarrow 0}\left(\frac{1}{t}-\frac{1}{t^{2}+t}\right) \) converges to 1. Thus, the value of the limit is 1.

Key Concepts

Common Denominator

When working with fractions, finding a common denominator is crucial to combining them into a single fraction. In our problem, we dealt with the fractions \( \frac{1}{t} \) and \( \frac{1}{t^2 + t} \). Combining these fractions requires a common platform where each fraction can be rewritten equivalently.

To achieve this, we need to find a common denominator for the fractions, which involves identifying the least common multiple of their denominators. For \( \frac{1}{t} \) and \( \frac{1}{t^2 + t} \), we recognized that the least common multiple is \( t(t + 1) \), since \( t^2 + t \) can be factored as \( t(t + 1) \).

Using this common denominator, we can rewrite our fractions equivalently. The fraction \( \frac{1}{t} \) becomes \( \frac{t+1}{t(t+1)} \), and \( \frac{1}{t^2 + t} \) simplifies directly to \( \frac{1}{t(t+1)} \). Doing so allows us to subtract the two, leading to the next step of simplification.

Simplification

Simplifying an expression is the process of reducing it to its most basic form. This is a valuable skill for evaluating limits and working with fractions in calculus.

In the expression \( \frac{1}{t} - \frac{1}{t^2 + t} = \frac{t}{t(t+1)} \), we are left with a common factor of \( t \) in both the numerator and the denominator. This is often the case with limit problems, where simplifying is crucial because it avoids indeterminate forms.

By canceling the \( t \) terms, the expression is reduced to \( \frac{1}{t+1} \). This form is significantly easier to work with, especially when evaluating the limit as \( t \rightarrow 0 \). Simplification not only makes calculations more straightforward but also helps to reveal the structure of the function that we're dealing with.

Evaluating Limits

Evaluating limits involves determining the value that a function approaches as the input approaches a particular point. Understanding limits is fundamental in calculus, as they underpin many concepts, including continuity and derivatives.

In this problem, after simplifying our expression to \( \frac{1}{t+1} \), we can easily substitute \( t = 0 \) to find the limit. This substitution gives \( \frac{1}{1} = 1 \).

• The importance of simplification becomes evident here, as it allows the direct substitution, giving a clear and direct answer.
• Without simplifying, you may face undefined conditions or complex calculations.

By evaluating the simplified form, we identify that the function approaches the value 1 as \( t \) gets closer to 0. Thus, mastering these three steps—finding a common denominator, simplifying, and evaluating limits—enables you to tackle more challenging calculus problems efficiently.