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Chapter 1: Problem 50

Find the limit (if it exists). $$ \lim _{x \rightarrow 2} \frac{2-x}{x^{2}-4} $$

Short Answer

Expert verified
The limit as \( x \) approaches 2 of the given function is \( \frac{1}{4} \)

Step by step solution

01

Factorization

Start by simplifying the function. The denominator can be factored as it's a difference of squares, \( a^{2} - b^{2} = (a-b)(a+b) \). So our expression becomes: \( \frac{2-x}{(x - 2)(x + 2)} \)
02

Simplifying the function

After factoring, notice that the numerator and one of the terms in the denominator cancel each other out. As the reverse of (x-2) is also (2-x), they cancel each other. The function simplifies to: \( \frac{1}{x + 2} \)
03

Find the limit

Now that we have simplified the function, we can substitute \(x = 2\). So, \( \lim_{x \rightarrow 2} \frac{1}{x + 2} = \frac{1}{2 + 2} \)

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Key Concepts

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

Factorization
Factorization is a mathematical technique where a complex expression is expressed as the product of simpler expressions. It can make some problems easier to solve, as seen in this exercise where finding the limit of a function becomes manageable through factorization. The key idea is to identify factors that can be multiplied together to get the original expression.

In our problem, the denominator was a difference of squares, which is factored by recognizing it as the form \( a^2 - b^2 \). We express it as the product \( (a - b)(a + b) \). It is essentially like breaking down a large object into smaller, more manageable pieces so that we can work with it effectively. By doing so, we were able to simplify the problem on our path to find the limit.
Simplifying Expressions
Simplifying expressions involves reducing a complex mathematical expression into its simplest form. This process often includes factorizing algebraic expressions, combining like terms, and cancelling out terms where possible.

In our exercise, simplifying the expression was crucial after factorization. We noticed that \(2 - x\) and \(x - 2\) are actually the same terms but with opposite signs, and hence they cancelled out. Simplification is like decluttering a room to make it easier to navigate. By reducing the complexity of the expression, the problem becomes clearer and more accessible, leading us directly to a simplified form \(\frac{1}{x + 2}\), setting the stage for the easy computation of the limit.
Difference of Squares
The difference of squares is a specific case of factorization which applies to expressions of the form \( a^2 - b^2 \). It's a fundamental algebraic identity where such an expression can be factored into \( (a - b)(a + b) \). This identity is particularly useful in calculus when finding limits, as it can transform a seemingly complex rational function into one that is much simpler.

For the given problem, recognizing the denominator as a difference of squares allowed us to factor it effectively. Knowing these patterns and how to apply them is akin to having a map in unfamiliar terrain. It can guide you to your destination much more quickly, and in mathematics, it leads you towards the solution.

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Most popular questions from this chapter

(a) Let \(f_{1}(x)\) and \(f_{2}(x)\) be continuous on the closed interval \([a, b]\). If \(f_{1}(a)f_{2}(b)\), prove that there exists \(c\) between \(a\) and \(b\) such that \(f_{1}(c)=f_{2}(c)\). (b) Show that there exists \(c\) in \(\left[0, \frac{\pi}{2}\right]\) such that \(\cos x=x\). Use a graphing utility to approximate \(c\) to three decimal places.

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(f(x)=g(x)\) for \(x \neq c\) and \(f(c) \neq g(c)\), then either \(f\) or \(g\) is not continuous at \(c\).

(a) Prove that if \(\lim _{x \rightarrow c}|f(x)|=0\), then \(\lim _{x \rightarrow c} f(x)=0\). (Note: This is the converse of Exercise \(110 .\) ) (b) Prove that if \(\lim _{x \rightarrow c} f(x)=L\), then \(\lim _{x \rightarrow c}|f(x)|=|L|\). [ Hint: Use the inequality \(\|f(x)|-| L\| \leq|f(x)-L| .]\)

Use a graphing utility to graph the function. Use the graph to determine any \(x\) -values at which the function is not continuous. $$ f(x)=\llbracket x \rrbracket-x $$

Find the limit (if it exists). If it does not exist, explain why. $$ \lim _{x \rightarrow-3^{-}} \frac{x}{\sqrt{x^{2}-9}} $$
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