For the rational expression $f(x) = \frac{6x^3 - 5x + 4}{2x^3 + 7x^2 - 9}$, rewrite $f(x)$ by dividing both numerator and denominator by the highest power of $x$ in the denominator to find an equivalent form that reveals its asymptotic limit as $x \to \infty$, and evaluate that asymptotic value $L$.

For the rational expression $f(x) = \frac{6x^3 - 5x + 4}{2x^3 + 7x^2 - 9}$ , rewrite $f(x)$ by dividing both numerator and denominator by the highest power of $x$ in the denominator to find an equivalent form that reveals its asymptotic limit as $x \to \infty$ , and evaluate that asymptotic value $L$ .

Final answer

Equivalent form: $\frac{6 - \frac{5}{x^2} + \frac{4}{x^3}}{2 + \frac{7}{x} - \frac{9}{x^3}}$ ; asymptotic limit: $L = 3$

Steps

Identify the highest power of $x$ in the denominator

To find the equivalent form and determine the horizontal asymptote, follow these steps:

In the rational expression: $f(x) = \frac{6x^3 - 5x + 4}{2x^3 + 7x^2 - 9}$ The highest power of $x$ in the denominator is $x^3$ .

Divide both numerator and denominator by $x^3$

Divide each term in the numerator and denominator by $x^3$ for $x \neq 0$ :

Numerator: $\frac{6x^3 - 5x + 4}{x^3} = \frac{6x^3}{x^3} - \frac{5x}{x^3} + \frac{4}{x^3} = 6 - \frac{5}{x^2} + \frac{4}{x^3}$
Denominator: $\frac{2x^3 + 7x^2 - 9}{x^3} = \frac{2x^3}{x^3} + \frac{7x^2}{x^3} - \frac{9}{x^3} = 2 + \frac{7}{x} - \frac{9}{x^3}$

Putting these together gives the equivalent form: $f(x) = \frac{6 - \frac{5}{x^2} + \frac{4}{x^3}}{2 + \frac{7}{x} - \frac{9}{x^3}}$

Evaluate the asymptotic limit as $x \to \infty$

As $x$ becomes arbitrarily large ( $x \to \infty$ ), any fraction of the form $\frac{c}{x^n}$ (where $c$ is a constant and $n > 0$ ) approaches $0$ : $\lim_{x \to \infty} \frac{5}{x^2} = 0, \quad \lim_{x \to \infty} \frac{4}{x^3} = 0, \quad \lim_{x \to \infty} \frac{7}{x} = 0, \quad \lim_{x \to \infty} \frac{9}{x^3} = 0$

Substitute these limiting values into the rewritten expression: $L = \lim_{x \to \infty} f(x) = \frac{6 - 0 + 0}{2 + 0 - 0} = \frac{6}{2} = 3$

Common Pitfall

A frequent mistake is dividing only the leading terms by $x^3$ while leaving constant or lower-degree terms unchanged. Be sure to divide every single term in both the numerator and the denominator by $x^3$ to keep the value of the rational expression unchanged.

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student misidentifies the highest power of x in the denominator as x² instead of x³, dividing each term by x² and incorrectly concluding that the limit is undefined or diverges.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
Highest power in denominator is $x^2$
Correct
The highest power of $x$ in the denominator $2x^3 + 7x^2 - 9$ is $x^3$ .
The denominator contains the term $2x^3$ , whose exponent is $3$ . Therefore, the highest power of $x$ in the denominator is $x^3$ , not $x^2$ .
Step 2
Divide numerator and denominator by $x^2 \implies f(x) = \frac{6x - \frac{5}{x} + \frac{4}{x^2}}{2x + 7 - \frac{9}{x^2}}$
Correct
f(x) = $\frac{\frac{6x^3 - 5x + 4}{x^3}}{\frac{2x^3 + 7x^2 - 9}{x^3}}$ = $\frac{6 - \frac{5}{x^2} + \frac{4}{x^3}}{2 + \frac{7}{x} - \frac{9}{x^3}}$
Because the highest power was identified as $x^2$ , dividing by $x^2$ leaves degree-1 terms ( $6x$ and $2x$ ) rather than constant leading coefficients.
Step 3
As $x \to \infty$ , $6x$ and $2x$ go to $\infty$
Correct
As $x \to \infty$ , $\frac{5}{x^2}$ $\to$ 0,; $\frac{4}{x^3}$ $\to$ 0,; $\frac{7}{x}$ $\to$ 0,; $\text{and }$ $\frac{9}{x^3}$ $\to$ 0.
This step evaluates the rewritten expression from Step 2, where linear terms remained because of dividing by $x^2$ instead of $x^3$ .
Step 4
L $\text{ does not exist}$
Correct
L = $\lim_{x $\to$ $\infty$ }$ f(x) = \frac{6 - 0 + 0}{2 + 0 - 0} = \frac{6}{2} = 3_
With the correct division by $x^3$ , the rational expression simplifies to constants plus reciprocal powers that vanish, yielding the finite limit $L = 3$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly divided each individual term in both the numerator and denominator by your chosen power.

You correctly identified that reciprocal powers of $x$ vanish as $x \to \infty$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Misidentified highest degree in denominator

You overlooked the leading term $2x^3$ in the denominator and selected $x^2$ as the highest power.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Failing to scan all terms in the denominator polynomial to find the maximum exponent before choosing the divisor.

REASONING PATH

Your ACEpath teacher’s suggested approach

Find Denominator Degree: Scan every term of the denominator polynomial and identify the term with the largest exponent.
Look at $2x^3 + 7x^2 - 9$ : the exponents are $3$ , $2$ , and $0$ . The maximum exponent is $3$ , so the highest power is $x^3$ .
Divide by the True Leading Power: Dividing by $x^{\text{deg}(\text{denom})}$ guarantees that the denominator approaches a non-zero constant as $x \to \infty$ .
Divide every term in both numerator and denominator by $x^3$ so that the leading terms reduce to the real constants $6$ and $2$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Identify the highest power of $x$ in the denominator $2x^3 + 7x^2 - 9$ , which is $x^3$ .
Divide every term in the numerator and denominator by $x^3$ : $f(x) = \frac{6 - \frac{5}{x^2} + \frac{4}{x^3}}{2 + \frac{7}{x} - \frac{9}{x^3}}$ .
Evaluate the limit as $x \to \infty$ : since $\frac{1}{x^n} \to 0$ for all $n > 0$ , the expression approaches $\frac{6 - 0 + 0}{2 + 0 - 0} = \frac{6}{2} = 3$ .

Final answer

Equivalent form: $\frac{6 - \frac{5}{x^2} + \frac{4}{x^3}}{2 + \frac{7}{x} - \frac{9}{x^3}}$ ; asymptotic limit: $L = 3$

NEXT STEPS

Final Takeaways from Your Teacher

Before dividing, write down the degree of each term in the denominator explicitly to confirm the maximum exponent.
Verify after dividing that the leading term of the denominator has become a non-zero constant, not a variable term.

OverviewACEpath Teacher's Diagnosis

02 The student incorrectly divides the middle term of the numerator by x³ as 5/x instead of 5/x², leading to an algebraic error in the rewritten equivalent form even though the downstream limit coincidentally converges.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
$\text{Highest power of }$ x $\text{ in denominator is }$ x^3
Step 2
f(x) = $\frac{6 - \frac{5}{x} + \frac{4}{x^3}}{2 + \frac{7}{x} - \frac{9}{x^3}}$
Correct
f(x) = $\frac{6 - \frac{5}{x^2} + \frac{4}{x^3}}{2 + \frac{7}{x} - \frac{9}{x^3}}$
When dividing $-5x$ by $x^3$ , subtracting the exponents gives $\frac{x^1}{x^3} = \frac{1}{x^{3-1}} = \frac{1}{x^2}$ , so the term must be $-\frac{5}{x^2}$ rather than $-\frac{5}{x}$ .
Step 3
$\text{As }$ x $\to$ $\infty$ , ; $\frac{5}{x}$ $\to$ 0, ; $\frac{4}{x^3}$ $\to$ 0, ; $\frac{7}{x}$ $\to$ 0, ; $\frac{9}{x^3}$ $\to$ 0
Correct
$\text{As }$ x $\to$ $\infty$ , ; $\frac{5}{x^2}$ $\to$ 0, ; $\frac{4}{x^3}$ $\to$ 0, ; $\frac{7}{x}$ $\to$ 0, ; $\frac{9}{x^3}$ $\to$ 0
Your reasoning that reciprocal power terms approach 0 as $x \to \infty$ is completely sound, though it followed your rewritten term $\frac{5}{x}$ instead of $\frac{5}{x^2}$ .
Step 4
L = $\frac{6}{2}$ = 3

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly identified the highest power of $x$ in the denominator as $x^3$ .

You correctly divided the denominator terms by $x^3$ to get $2 + \frac{7}{x} - \frac{9}{x^3}$ .

You applied the asymptotic limit property that $\lim_{x \to \infty} \frac{c}{x^n} = 0$ for $n > 0$ correctly.

You calculated the final asymptotic limit $L = 3$ accurately.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Exponent subtraction error during rational simplification

You reduced $\frac{5x}{x^3}$ to $\frac{5}{x}$ instead of $\frac{5}{x^2}$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Applying quotient rule of exponents carelessly on linear terms: forgetting that $x = x^1$ , which yields $x^{1 - 3} = x^{-2} = \frac{1}{x^2}$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Exponent subtraction check: For any positive integers $m$ and $n$ with $n > m$ , $\frac{x^m}{x^n} = \frac{1}{x^{n-m}}$ .
Always write explicit exponents when simplifying terms like $\frac{x^1}{x^3} = \frac{1}{x^{3-1}} = \frac{1}{x^2}$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Divide each individual term in the numerator by $x^3$ : $\frac{6x^3}{x^3} - \frac{5x^1}{x^3} + \frac{4}{x^3} = 6 - \frac{5}{x^2} + \frac{4}{x^3}$ .
Divide each individual term in the denominator by $x^3$ : $\frac{2x^3}{x^3} + \frac{7x^2}{x^3} - \frac{9}{x^3} = 2 + \frac{7}{x} - \frac{9}{x^3}$ .
Combine both to express the exact equivalent form: $f(x) = \frac{6 - \frac{5}{x^2} + \frac{4}{x^3}}{2 + \frac{7}{x} - \frac{9}{x^3}}$ .

Final answer

Equivalent form: $\frac{6 - \frac{5}{x^2} + \frac{4}{x^3}}{2 + \frac{7}{x} - \frac{9}{x^3}}$ ; asymptotic limit: $L = 3$

NEXT STEPS

Final Takeaways from Your Teacher

Double-check that you treat $x$ as $x^1$ when dividing powers of $x$ .
Verify each term's simplified power by adding exponents back: $\frac{1}{x^2} \cdot x^3 = x$ , whereas $\frac{1}{x} \cdot x^3 = x^2 \neq x$ .

OverviewACEpath Teacher's Diagnosis

03 The student evaluates the reciprocal power terms as approaching 1 instead of 0 as x approaches infinity, resulting in an incorrect limit value.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
f(x) = $\frac{6 - \frac{5}{x^2} + \frac{4}{x^3}}{2 + \frac{7}{x} - \frac{9}{x^3}}$
Step 2
$\text{As }$ x $\to$ $\infty$ , $\text{ all }$ $\frac{1}{x}$ $\text{ terms equal }$ 1
Correct
$\lim_{x $\to$ $\infty$ }$ \frac{1}{x^n} = 0 \text{ for any } n > 0_
As $x$ grows without bound, dividing a constant by an increasingly large number yields a value arbitrarily close to $0$ , not $1$ .
Step 3
L = $\frac{6 - 5 + 4}{2 + 7 - 9}$ $\implies$ L = $\frac{5}{0}$ = $\text{undefined}$
Correct
L = $\frac{6 - 0 + 0}{2 + 0 - 0}$ = $\frac{6}{2}$ = 3
Substituting $1$ for the reciprocal terms carried the previous misconception forward, producing a zero denominator.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly divided every term in the rational expression by the highest denominator power $x^3$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Reciprocal limit misconception

none

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You treated reciprocal terms $\frac{c}{x^n}$ as tending to $1$ rather than $0$ when $x \to \infty$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Reciprocal Limit Property: For any real constant $c$ and positive power $n > 0$ , $\lim_{x \to \infty} \frac{c}{x^n} = 0$ .
Whenever evaluating a limit at infinity, replace each term with $x$ in its denominator with $0$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Evaluate the limit of each reciprocal term as $x \to \infty$ : $\lim_{x \to \infty} \frac{5}{x^2} = 0$ , $\lim_{x \to \infty} \frac{4}{x^3} = 0$ , $\lim_{x \to \infty} \frac{7}{x} = 0$ , and $\lim_{x \to \infty} \frac{9}{x^3} = 0$ .
Substitute these zero limits into the equivalent expression: $L = \frac{6 - 0 + 0}{2 + 0 - 0} = \frac{6}{2} = 3$ .

Final answer

Equivalent form: $\frac{6 - \frac{5}{x^2} + \frac{4}{x^3}}{2 + \frac{7}{x} - \frac{9}{x^3}}$ ; asymptotic limit: $L = 3$

NEXT STEPS

Final Takeaways from Your Teacher

Review why $\frac{1}{x} \to 0$ as $x \to \infty$ by testing large values such as $x = 100$ and $x = 10{,}000$ .
Practice computing limits at infinity for rational functions by identifying and zeroing out all terms of the form $\frac{c}{x^n}$ .

Lock it in

Question

How do you determine the power of $x$ to divide by when finding the asymptotic limit of a rational function as $x \to \infty$ ?

Answer

Identify the highest exponent among all terms in the denominator, and divide every term in both numerator and denominator by $x$ raised to that power.

Resources

Answer and steps

Final answer

Steps

Common Pitfall

Typical errors

01 The student misidentifies the highest power of x in the denominator as x² instead of x³, dividing each term by x² and incorrectly concluding that the limit is undefined or diverges.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

02 The student incorrectly divides the middle term of the numerator by x³ as 5/x instead of 5/x², leading to an algebraic error in the rewritten equivalent form even though the downstream limit coincidentally converges.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

03 The student evaluates the reciprocal power terms as approaching 1 instead of 0 as x approaches infinity, resulting in an incorrect limit value.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

Lock it in

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