Problem 25
Question
Evaluate the given limit. $$ \lim _{x \rightarrow \infty} \frac{x^{3}+2 x^{2}+1}{x-5} $$
Step-by-Step Solution
Verified Answer
The limit is \( \infty \).
1Step 1: Identify the Dominant Terms
In the expression \( \frac{x^3+2x^2+1}{x-5} \), the dominant terms as \( x \rightarrow \infty \) are \( x^3 \) in the numerator and \( x \) in the denominator. These terms grow faster than the others, so they will determine the behavior of the limit.
2Step 2: Simplify the Expression
To simplify \( \frac{x^3+2x^2+1}{x-5} \), divide every term in the numerator and the denominator by \( x \), which is the highest power in the denominator. This gives us \( \frac{x^2 + \frac{2}{x}x + \frac{1}{x^2}}{1 - \frac{5}{x}} \).
3Step 3: Evaluate the Limit
As \( x \rightarrow \infty \), the terms \( \frac{2}{x} \), \( \frac{1}{x^2} \), and \( \frac{5}{x} \) all approach 0. Therefore, the expression simplifies to \( \frac{x^2 + 0 + 0}{1 - 0} = x^2 \).
4Step 4: Determine the Limit's Behavior
The result \( x^2 \) indicates that the limit grows without bound as \( x \rightarrow \infty \). Therefore, the limit does not approach a finite number.
Key Concepts
Dominant TermsSimplifying ExpressionsBehavior of Limits
Dominant Terms
When working with limits at infinity, identifying dominant terms becomes crucial. We use dominant terms to predict how functions behave as variables grow very large. In the expression \( \frac{x^3 + 2x^2 + 1}{x - 5} \), we are interested in which terms in the numerator and denominator increase at the fastest rate as \( x \to \infty \).
In our example:
In our example:
- Numerator: The term \( x^3 \) grows much faster than the terms \( 2x^2 \) and \( 1 \) as \( x \) increases.
- Denominator: The term \( x \) is the only term in the denominator, making it dominant by default.
Simplifying Expressions
Once dominant terms are identified, the next step is simplifying the expression to make it easier to evaluate the limit. We do this by dividing each term by the highest power of \( x \) present in the denominator.
In the given expression \( \frac{x^3 + 2x^2 + 1}{x - 5} \), we divide each term by \( x \):
In the given expression \( \frac{x^3 + 2x^2 + 1}{x - 5} \), we divide each term by \( x \):
- Numerator becomes \( x^2 + \frac{2}{x}x + \frac{1}{x^2} \).
- Denominator becomes \( 1 - \frac{5}{x} \).
Behavior of Limits
The ultimate goal is understanding how an expression behaves as \( x \to \infty \). After simplifying the expression in our problem, it becomes \( \frac{x^2 + 0 + 0}{1 - 0} \), which simplifies to \( x^2 \).
As \( x \) increases, \( x^2 \) expands without bounds, indicating that the limit of the expression is infinite.
Key insights about behavior of limits:
As \( x \) increases, \( x^2 \) expands without bounds, indicating that the limit of the expression is infinite.
Key insights about behavior of limits:
- If, after simplification, the expression tends to a constant, the limit is that constant.
- If it contains terms that grow endlessly, like \( x^2 \), the limit is infinite.
- If it oscillates without settling towards a single value, the limit does not exist.
Other exercises in this chapter
Problem 24
Evaluate the given limit. $$ \lim _{x \rightarrow 3} 4^{x^{3}-8 x} $$
View solution Problem 24
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 25
Give the intervals on which the given function is continuous. $$ g(x)=\sqrt{4-x^{2}} $$
View solution Problem 25
Approximate the limit numerically: \(\lim _{x \rightarrow 0.4} \frac{x^{2}-4.4 x+1.6}{x^{2}-0.4 x}\).
View solution