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Chapter 2: Problem 18

Find the limits if they exist. An \(\varepsilon-\delta\) test is not required. $$ \lim _{x \rightarrow 5} \frac{x^{2}-25}{x-5} $$

Short Answer

Expert verified
The limit is 10.

Step by step solution

01

Substitute the Limit Value

Begin by directly substituting the limit value, \( x = 5 \), into the function to see if the limit can be evaluated directly.\[\lim_{x \to 5} \frac{x^2 - 25}{x - 5} = \frac{5^2 - 25}{5 - 5} = \frac{0}{0}\]Since the result is \(\frac{0}{0}\), this is an indeterminate form, and further analysis is needed.
02

Factor the Numerator

To resolve the indeterminate form, factor the numerator \( x^2 - 25 \) using the difference of squares formula:\[x^2 - 25 = (x - 5)(x + 5)\]
03

Simplify the Expression

Rewrite the limit expression by substituting the factored form of the numerator:\[\lim_{x \to 5} \frac{(x - 5)(x + 5)}{x - 5}\]Then, cancel the common factor \( x - 5 \) from the numerator and denominator:\[\lim_{x \to 5} (x + 5)\]
04

Evaluate the Limit

Now that the expression is simplified, substitute \( x = 5 \) directly into \( x + 5 \):\[\lim_{x \to 5} (x + 5) = 5 + 5 = 10\]

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Key Concepts

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

Indeterminate Forms
In calculus, an indeterminate form like \(\frac{0}{0}\) occurs when both the numerator and the denominator of a function approach zero as \(x\) approaches some limit. This means that directly substituting the value doesn't give us useful information about the limit's behavior. It's like trying to divide nothing by nothing, which doesn't allow us to determine anything specific about the result directly. Indeterminate forms indicate that we need to use additional algebraic techniques to further investigate and resolve the expression, such as factoring or other methods, to determine the true value of the limit.
Difference of Squares
The difference of squares is a specific formula that helps in factoring polynomial expressions and is used quite often in calculus to simplify complex expressions. This identity is represented as \(a^2 - b^2 = (a-b)(a+b)\).

For example, in the limit problem given, \(x^2 - 25\) can be viewed as a difference of squares. Here, \(a\) would be \(x\) and \(b\) would be 5 because 25 is \(5^2\). Thus, the expression \(x^2 - 25\) can be factored into \((x-5)(x+5)\).

This technique of factoring helps to simplify expressions, making it easier to carry out further operations like cancellation and evaluation, as shown in the steps provided in the solution.
Simplifying Expressions
Simplifying expressions is a key skill in calculus. After identifying and using mathematical identities, like the difference of squares, we often end up with expressions that can be reduced further.

In our problem, after factoring \(x^2 - 25\) as \((x-5)(x+5)\), the expression \(\frac{(x-5)(x+5)}{x-5}\) includes \(x-5\) in both the numerator and the denominator.

These terms can be cancelled out because \(x\) does not equal 5 during the simplification process (this is why we're analyzing limits).
  • This cancellation helps in removing the indeterminate form initially present.
  • We are left with the much simpler \(x+5\), which can be directly evaluated once the limit is applied.
This step is crucial for determining the final solution of a limit problem efficiently.

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Most popular questions from this chapter

From an income of \(x=\$ 10,000\) you spend \(y=\$ 1200\) on your car. If \(E=\frac{1}{2},\) what fraction of your next dollar will be spent on the car? Compare \(d y / d x\) (marginal) with \(y / x\) (average).

Name two functions with \(d f / d x=1 / x^{2}\).

What is \(a_{4}\) and what is the limit \(L\) ? After which \(N\) is \(\left|a_{n}-L\right|<\frac{1}{10} ?\) (Calculator allowed) (a) \(-1,+\frac{1}{2},-\frac{1}{3}, \ldots\) (b) \(\frac{1}{2}, \frac{1}{2}+\frac{1}{4}, \frac{1}{2}+\frac{1}{4}+\frac{1}{6}, \ldots\) (c) \(\frac{1}{2}, \frac{2}{4}, \frac{3}{8}, \ldots a_{n}=n / 2^{n}\) (d) \(1.1,1.11,1.111, \ldots\) (e) \(a_{n}=\sqrt[n]{n}\) (f) \(a_{n}=\sqrt{n^{2}+n}-n\) (g) \(1+1,\left(1+\frac{1}{2}\right)^{2},\left(1+\frac{1}{3}\right)^{3}, \ldots\)

\(f(x)\) is any continuous function with \(f(0)=f(1) .\) (a) Draw a typical \(f(x)\). Mark where \(f(x)=f\left(x+\frac{1}{2}\right)\). (b) Explain why \(g(x)=f\left(x+\frac{1}{2}\right)-f(x)\) has \(g\left(\frac{1}{2}\right)=-g(0)\). (c) Deduce from (b) that (a) is always possible: There must be a point where \(g(x)=0\) and \(f(x)=f\left(x+\frac{1}{2}\right)\)

Show by example that these statements are false: (a) If \(a_{n} \rightarrow L\) and \(b_{n} \rightarrow L\) then \(a_{n} / b_{n} \rightarrow 1\) (b) \(a_{n} \rightarrow L\) if and only if \(a_{n}^{2} \rightarrow L^{2}\) (c) If \(a_{n}<0\) and \(a_{n} \rightarrow L\) then \(L<0\) (d) If infinitely many \(a_{n}\) 's are inside every strip around zero then \(a_{n} \rightarrow 0\).
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