Problem 12 - Calculus Early Transcendentals

Question

\(10-12\) Use the definition of continuity and the properties of limits to show that the function is continuous at the given number a. \(h(t)=\frac{2 t-3 t^{2}}{1+t^{3}}, \quad a=1\)

Step-by-Step Solution

Answer

The function is continuous at \(t = 1\) because \(\lim_{t \to 1} h(t) = h(1) = -\frac{1}{2}\).

Step 1: Understanding Continuity

A function is continuous at a point \(a\) if \(\lim_{t \to a} h(t) = h(a)\). We need to calculate both \(h(1)\) and \(\lim_{t \to 1} h(t)\), and show that they are equal.

Step 2: Computing \(h(1)\)

Substitute \(t = 1\) into the function \(h(t)\): \[h(1) = \frac{2(1) - 3(1)^2}{1 + (1)^3} = \frac{2 - 3}{1 + 1} = \frac{-1}{2}\] So, \(h(1) = -\frac{1}{2}\).

Step 3: Calculating the Limit \(\lim_{t \to 1} h(t)\)

Compute the limit: \[\lim_{t \to 1} \frac{2t - 3t^2}{1 + t^3}\] To do this, merge terms and simplify if needed. The function itself is a rational function and, at \(t = 1\), the expression does not become undefined, so:\[ = \frac{2(1) - 3(1)^2}{1 + 1^3} = \frac{-1}{2}\].

Step 4: Comparing Limit and Function Value

Now, compare \(\lim_{t \to 1} h(t)\) and \(h(1)\):\[\lim_{t \to 1} h(t) = -\frac{1}{2} = h(1)\].Both are equal, confirming the function is continuous at \(t = 1\).

Key Concepts

Definition of Continuity

Continuity is a fundamental concept in calculus that helps us understand the behavior of functions. To say that a function is continuous at a point means there are no breaks, jumps, or holes at that point in the graph of the function. Specifically, a function \( f(x) \) is continuous at a point \( a \) if three conditions are met:

• The function \( f(x) \) is defined at \( a \).
• The limit \( \lim_{x \to a} f(x) \) exists.
• The limit equals the function's value at \( a \): \( \lim_{x \to a} f(x) = f(a) \).

In our specific case with \(h(t)\), we've confirmed it is continuous at \( t = 1 \) by finding both \( h(1) \) and \( \lim_{t \to 1} h(t) \), and checking they are equal.

Properties of Limits

Limits play a crucial role in determining continuity and overall function behavior. The limit describes what the function approaches as the input approaches some value. Several properties of limits help simplify the computation:

• Addition/Subtraction: \( \lim_{x \to a} [f(x) \pm g(x)] = \lim_{x \to a} f(x) \pm \lim_{x \to a} g(x) \)
• Multiplication: \( \lim_{x \to a} [f(x) \cdot g(x)] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x) \)
• Division: \( \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} \) if \( \lim_{x \to a} g(x) eq 0 \)

By applying these properties, we evaluated the limit \( \lim_{t \to 1} \frac{2t - 3t^2}{1 + t^3} \) without complications because the denominator is not zero at \( t = 1 \). This confirms the functioning of these limit properties when assessing continuity.

Rational Functions

Rational functions are quotients of polynomials, like \( h(t) = \frac{2t - 3t^2}{1 + t^3} \). These kinds of functions have unique characteristics due to being composed of polynomials:

• They are continuous wherever the denominator is non-zero, which ensures the function is defined.
• Rational functions can be simplified by cancelling common factors, given they don't turn the denominator into zero.
• The limit of a rational function as \( t \) approaches a value usually involves direct substitution unless the function is undefined at the point.

In the example provided, since the denominator \(1 + t^3\) is not zero at \( t = 1 \), the function \( h(t) \) remains continuous there. Understanding these features makes evaluating and proving the continuity straightforward, as was shown in the solution by directly substituting \( t = 1 \) into the function.