Problem 22
Question
Evaluate the given limit. $$ \lim _{x \rightarrow 1} \frac{2 x-2}{x+4} $$
Step-by-Step Solution
Verified Answer
The limit is 0.
1Step 1: Substitute the limit value into the expression
To evaluate the limit, start by substituting the value that x is approaching, which is 1, directly into the expression. Substitute x = 1 into \( \frac{2x-2}{x+4} \), resulting in \( \frac{2(1)-2}{1+4} = \frac{0}{5} \).
2Step 2: Simplify the expression
Since direct substitution resulted in a determinate form, simplify the expression if possible. In this case, \( \frac{0}{5} \) is already simplified as 0.
3Step 3: Conclude the limit evaluation
Since the resulting expression is \( \frac{0}{5} \), the limit can be concluded as 0.
Key Concepts
Direct SubstitutionDeterminate FormSimplifying Expressions
Direct Substitution
When evaluating the limit of a function as it approaches a certain value, one of the first methods to consider is direct substitution. This technique involves directly inserting the point into the limit function to see if it results in a simplifiable expression.
- Start by identifying the point that the variable is approaching, which is often given directly in the exercise. In our case, it is when \(x\) approaches 1.
- Substitute this value into the function. Here, you substitute \(x = 1\) into \( \frac{2x-2}{x+4} \).
This calculation gives \( \frac{2(1)-2}{1+4} = \frac{0}{5} \).
Determinate Form
The term "determinate form" refers to the outcome of a substitution in a limit evaluation, where the expression results in a specific, concrete value. This is opposite of an indeterminate form, which is one of the unreadable expressions like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).
- If, after substitution, the expression resolves to a clear, finite value, it is a determinate form.
In this example: \(\frac{0}{5}\) is a determinable end solution, because the "0" implies a result to be exact and does not require further calculation. - Determinate forms mean that additional techniques like factoring, rationalizing, or otherwise manipulating the function might not be necessary, as there are no uncertainties left in the final result.
Simplifying Expressions
Simplification is a key principle in solving mathematical problems, including those involving limits. Once you've substituted and found a result, it’s always good practice to check if it can be simplified further.
- First, ensure that the numerator and denominator cannot be reduced further.
For the expression \(\frac{0}{5}\), the numerator is already zero, which makes the entire expression zero. - Simplification ensures that the expression is in its most basic form, revealing insights into the function's behavior.
In many cases, like this example, thoughtful simplification confirms the limit as \(0\).
Other exercises in this chapter
Problem 22
Determine if \(f\) is continuous at the indicated values. If not, explain why. \(f(x)=\left\\{\begin{array}{cl}\frac{x^{2}-64}{x^{2}-11 x+24} & x \neq 8 \\ 5 &
View solution Problem 22
Evaluate the limit: \(\lim _{x \rightarrow-1} \frac{x^{2}+5 x+4}{x^{2}-3 x-4}\).
View solution Problem 22
A function \(f\) and \(a\) value \(a\) are given. Approximate the limit of the difference quotient, \(\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h},\) using \(h
View solution Problem 23
Identify the horizontal and vertical asymptotes, if any, of the given function. $$ f(x)=\frac{x^{2}-9}{9 x+27} $$
View solution