Problem 12 - Calculus: Early Transcendentals

Question

Use the definition of continuity and the properties of limits to show that the function is continuous at the given number \( a \). \( g(t) = \frac{t^2 + 5t}{2t + 1}, \hspace{5mm} a = 2 \)

Step-by-Step Solution

Answer

The function \( g(t) \) is continuous at \( t = 2 \) because \( \lim_{t \to 2} g(t) = g(2) = 2.8 \).

Step 1: Understand the Problem

We need to prove that the function \( g(t) = \frac{t^2 + 5t}{2t + 1} \) is continuous at \( t = 2 \). To do this, we must show that \( \lim_{t \to 2} g(t) = g(2) \).

Step 2: Evaluate the Function at t = 2

First, compute \( g(2) \) by substituting \( t = 2 \) into the function:\[g(2) = \frac{2^2 + 5 \cdot 2}{2 \cdot 2 + 1} = \frac{4 + 10}{4 + 1} = \frac{14}{5} = 2.8\]

Step 3: Find the Limit as t Approaches 2

Next, find \( \lim_{t \to 2} g(t) \). Substitute and simplify:\[g(t) = \frac{t^2 + 5t}{2t + 1}\]The limit can be directly computed by substituting \( t = 2 \):\[\lim_{t \to 2} g(t) = \lim_{t \to 2} \frac{t^2 + 5t}{2t + 1} = \frac{2^2 + 5 \cdot 2}{2 \cdot 2 + 1} = \frac{14}{5} = 2.8\]

Step 4: Compare Limit and Function Value

Now verify whether \( \lim_{t \to 2} g(t) = g(2) \):Both \( \lim_{t \to 2} g(t) = 2.8 \) and \( g(2) = 2.8 \). Therefore, the function is continuous at \( t = 2 \).

Key Concepts

Limits in Calculus

Limits in calculus are essential because they help us understand the behavior of functions as they approach a specific point. When computing a limit, we want to see how a function behaves as its input gets closer to some value. The limit provides insight into what the output of the function would be, without necessarily plugging in the value directly into the function.

Consider the original problem, where we were asked to determine if the function is continuous at a specific point using limits:

• We need to compute the limit of the function as the variable approaches the point of interest.
• For the function given, this means finding \( \lim_{t \to 2} g(t) \).
• The limit exists if we can approach the point from both sides, and the function approaches a single defined value.

Evaluating a limit can sometimes involve algebraic manipulation, but in many simple cases, substitute the point directly to see if the output is defined and finite. The function \( g(t) = \frac{t^2 + 5t}{2t + 1} \) was straightforward to evaluate by direct substitution in this scenario.

Evaluating Functions

Evaluating functions involves determining the output value for a given input. It's about applying the function's rules to specific inputs to find what they yield.

To evaluate the function at a point, consider the following steps:

• Replace the variable in the function with the specific number given.
• Execute the operations as per arithmetic rules to find the answer.

In the problem, we evaluated \( g(t) = \frac{t^2 + 5t}{2t + 1} \) at \( t = 2 \):

• First, substitute \( t = 2 \) into the function: \( g(2) = \frac{2^2 + 5 \cdot 2}{2 \cdot 2 + 1} \).
• Calculate: \( \frac{4 + 10}{4 + 1} = \frac{14}{5} \).
• The resulting value is \( 2.8 \).

By performing these steps, we find the function value at that point, which is necessary to determine continuity.

Continuity at a Point

Continuity at a point in calculus means that a function does not "jump" or have breaks at that point. For a function to be continuous at a point, three conditions must be met:

• The function is defined at the point: \( f(a) \) exists.
• The limit of the function as it approaches the point exists: \( \lim_{x \to a} f(x) \) is finite and exists.
• The limit of the function equals the function value: \( \lim_{x \to a} f(x) = f(a) \).

In the given exercise, we proved continuity at \( t = 2 \) by showing that:

• \( g(2) \) exists and is \( 2.8 \).
• The computed limit \( \lim_{t \to 2} g(t) \) is also \( 2.8 \).
• Both conditions are equal, confirming that the function is continuous at \( t = 2 \).

This is a fundamental property in calculus, ensuring the function's smoothness and predictability at a point.