Problem 56 - Thomas Calculus

Question

Find the limits in Exercises \(53-58\). $$\lim \frac{x^{2}-1}{2 x+4} \mathrm{as}$$ \begin{equation} \begin{array}{ll}{\text { a. }} & {x \rightarrow-2^{+}} & {\text { b. }} & {x \rightarrow-2^{-}} \\ {\text { c. }} & {x \rightarrow 1^{+}} & {\text { d. }} & {x \rightarrow 0^{-}}\end{array} \end{equation}

Step-by-Step Solution

Answer

a. \(+\infty\); b. \(-\infty\); c. 0; d. \(-\frac{1}{4}\).

Step 1: Analyzing Part a

To find \( \lim_{x \to -2^+} \frac{x^2-1}{2x+4} \), we start by noticing that \( 2x + 4 \) simplifies to \( 2(x + 2) \). When \( x \) approaches \(-2^+ \), \( x + 2 \) approaches 0 from the positive side. Thus the denominator approaches zero while the numerator \( x^2 - 1 \) equals \( 3 \) when \( x = -2 \). Therefore, the limit is infinity, \( +\infty \).

Step 2: Analyzing Part b

For part b, \( \lim_{x \to -2^-} \frac{x^2-1}{2x+4} \), the expression \( 2x + 4 \) in the denominator will approach 0 from the negative side as \( x \) approaches \(-2^- \). The numerator \( x^2 - 1 \) approaches 3. Hence, the limit is negative infinity, \( -\infty \).

Step 3: Analyzing Part c

For \( \lim_{x \to 1^+} \frac{x^2-1}{2x+4} \), substitute \( x = 1 \) in the expression. We have \( \frac{1^2 - 1}{2(1) + 4} = \frac{0}{6} = 0 \). As the expression is continuous around \( x = 1 \), the limit is 0.

Step 4: Analyzing Part d

For part d, examining \( \lim_{x \to 0^-} \frac{x^2-1}{2x+4} \), \( x \) approaches 0 from the negative side. Substitute \( x = 0 \) to get \( \frac{0^2 - 1}{0 + 4} = -\frac{1}{4} \). As this point is continuous, the limit is \(-\frac{1}{4}\).

Key Concepts

Continuous Functions

Continuous functions are essential in calculus because they help us predict behavior at certain points. A function is continuous if there are no jumps, breaks, or holes in its graph.
Think of a continuous function as a seamless trail where you can draw without lifting your pencil. For the function \(\frac{x^2 - 1}{2x + 4}\), continuity simplifies calculations considerably.
At any point where the expression does not result in a division by zero, the function is typically continuous.

• Continuous at a point means you can substitute the x-value directly to find the limit.
• For example, as \(x\) approaches 1 (in part c), the function is continuous, allowing us to plug 1 into the expression to find the limit exactly.

Understanding continuity helps solve many limit problems straightforwardly.

Infinity in Limits

Infinity in limits arises when a function increases or decreases without bound as it approaches a specific point.
Consider \(\lim_{x \to -2^+} \frac{x^2 - 1}{2x + 4}\) from the exercise. As \(x\) nears \(-2\) from the right, the denominator \(2x + 4\) approaches zero. However, it does so positively, causing the overall value of the fraction to skyrocket toward infinity.

• When limits tend to infinity, it typically means the output grows beyond any fixed number.
• The concept of infinity helps us understand asymptotic behaviors—a behavior where a function shoots sharply upward or downward as it closes in on a vertical line in its graph.

Recognizing scenarios of infinity is crucial for analyzing functions that diverge significantly.

Approaching Zero in Limits

Limits involving approaching zero are vital, especially for fractions that could lead to infinite limits or undefined forms. When approaching zero in the denominator, interesting behaviors emerge in limits.
For example, \(\lim_{x \to -2^-} \frac{x^2 - 1}{2x + 4}\) in the exercise shows how approaching zero from the negative side significantly affects the limit's result.

• If the denominator nears zero, assess whether it approaches from the positive or negative side, as in part b.
• Approaching zero can sometimes lead to finite limits, depending on the numerator's behavior.
• Even approaching zero in the numerator, seen in part c, can simplify to a zero limit.

Understanding how expressions behave as they approach zero is a fundamental skill in mastering limits.