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Chapter 2: Problem 3

Evaluate the limit.\(\lim _{t \rightarrow 2^{+}} \frac{t+2}{t^{2}-4}\)

Short Answer

Expert verified
\text{The limit is }\boldsymbol{\text{positive infinity.}}

Step by step solution

01

Understand the limit expression

We are given the limit as t approaches 2 from the right of the fraction \(\frac{t+2}{t^{2}-4}\). Our goal is to evaluate this limit.
02

Factor the denominator

First, factor the denominator \(t^{2} - 4\). Notice that \(t^{2} - 4\) can be factored as a difference of squares: \((t - 2)(t + 2)\). Therefore, the expression becomes \(\frac{t+2}{(t-2)(t+2)}\).
03

Simplify the expression

After factoring, the expression simplifies because \(t + 2\) in the numerator and denominator cancels out, yielding \(\frac{1}{t - 2}\).
04

Evaluate the limit

Now we evaluate the limit of \(\frac{1}{t - 2}\) as t approaches 2 from the right (2鈦�). As t gets closer to 2 from the right-hand side, \(t - 2\) gets closer to 0 but remains positive. Therefore, the fraction \(\frac{1}{t - 2}\) grows very large and positive. Hence, \(\frac{1}{t - 2}\) approaches positive infinity.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Right-hand limit
A right-hand limit is the value that a function approaches as the variable gets close to a point from the right side. For example, \[\frac{1}{t - 2}\] as \(t \to 2^+\). This means we're looking at values of \(t\) that are slightly greater than 2. When evaluating right-hand limits, always consider the direction from which you're approaching the point. In this case, as \(t \to 2^+\), the denominator \(t - 2 \) gets smaller but remains positive, which makes the entire fraction very large.
Factoring polynomials
Factoring polynomials simplifies complex expressions. Here鈥檚 how we factored the denominator \(t^2 - 4\). This expression is a difference of squares, which can be written as \(a^2 - b^2 = (a - b)(a + b)\). Applying this, \(t^2 - 4 = (t - 2)(t + 2)\). This transforms our original function \[\frac{t+2}{t^2-4} = \frac{t+2}{(t-2)(t+2)}\]. Factoring helps simplify the expression further.
Simplifying rational expressions
Simplifying rational expressions involves canceling out common factors between the numerator and denominator. After factoring the denominator, our expression was \[\frac{t + 2}{(t - 2)(t + 2)}\]. The \(t + 2\) terms in the numerator and denominator cancel each other out, simplifying our expression to \[\frac{1}{t - 2}\]. Simplification is crucial for making limits and other calculations easier to handle.
Infinite limits
Infinite limits occur when the value of a function increases or decreases without bound as the variable approaches a certain point. In our example, as \(t \to 2^+\), \(t - 2\) is a very small positive number, making \[\frac{1}{t - 2}\] a very large positive number. Therefore, the limit \(\frac{1}{t - 2}\) as \(t \to 2^+\) is positive infinity. This is an infinite limit because the function doesn't approach a finite number, it grows indefinitely larger.

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Most popular questions from this chapter

Establish the limit by using Definition 2.1.1; that is, for any \(\epsilon>0\), find a \(\delta>0\), such that \(|f(x)-L|<\epsilon\) whenever \(0<|x-a|<\delta$$\lim _{x \rightarrow 4} \frac{\sqrt{x}-2}{x-4}=\frac{1}{4}\)

Draw a sketch of the graph of the function; then by observing where there are breaks in the graph, determine the values of the independent variable at which the function is discontinuous and show why Definition 2.5.1 is not satisfied at each discontinuity.\(g(x)= \begin{cases}2 x+3 & \text { if } x \leq 1 \\\ 8-3 x & \text { if } 1

Find the value of the limit and when applicable indicate the limit theorems being used.\(\lim _{y \rightarrow-3} \sqrt{\frac{y^{2}-9}{2 y^{2}+7 y+3}}\)

Prove that if \(f\) is continuous at \(a\) and \(g\) is discontinuous at \(a\), then \(f+g\) is discontinuous at \(a\).

Prove that the function is discontinuous at the number \(a\). Then determine if the discontinuity is removable or essential. If the discontinuity is removable, define \(f(a)\) so that the discontinuity is removed.\(f(x)=\frac{\sqrt{2+\sqrt[3]{x}}-2}{x-8} ; a=8\)
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