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Evaluate the asymptotic behavior of the rational power function g(t)=5t−2/3+8t−1/3+123t−2/3−4t−1/3+2g(t) = \frac{5 t^{-2/3} + 8 t^{-1/3} + 12}{3 t^{-2/3} - 4 t^{-1/3} + 2} in two extreme parameter limits: (a) as t→∞t \to \infty, and (b) as t→0+t \to 0^+. Express your answer as the ordered pair of limits (lim⁡t→∞g(t),lim⁡t→0+g(t))\left( \lim_{t \to \infty} g(t), \lim_{t \to 0^+} g(t) \right).

Answer and steps

Final answer

(6,53)\left(6, \frac{5}{3}\right)

Steps

01

Limit as t→∞t \to \infty

To evaluate the behavior of g(t)=5t−2/3+8t−1/3+123t−2/3−4t−1/3+2g(t) = \frac{5 t^{-2/3} + 8 t^{-1/3} + 12}{3 t^{-2/3} - 4 t^{-1/3} + 2} in the two limits, analyze which terms dominate as t→∞t \to \infty and as t→0+t \to 0^+.

Recall the negative exponent rule: t−p=1tpt^{-p} = \frac{1}{t^p}. For any positive power p>0p > 0, as t→∞t \to \infty, the denominator grows without bound, so: lim⁡t→∞1tp=0\lim_{t \to \infty} \frac{1}{t^p} = 0

Applying this rule to the fractional powers:

  • lim⁡t→∞t−2/3=lim⁡t→∞1t2/3=0\lim_{t \to \infty} t^{-2/3} = \lim_{t \to \infty} \frac{1}{t^{2/3}} = 0
  • lim⁡t→∞t−1/3=lim⁡t→∞1t1/3=0\lim_{t \to \infty} t^{-1/3} = \lim_{t \to \infty} \frac{1}{t^{1/3}} = 0

Substitute these limiting values directly into the expression: lim⁡t→∞g(t)=5(0)+8(0)+123(0)−4(0)+2=122=6\lim_{t \to \infty} g(t) = \frac{5(0) + 8(0) + 12}{3(0) - 4(0) + 2} = \frac{12}{2} = 6

02

Limit as t→0+t \to 0^+

As t→0+t \to 0^+, negative powers grow unbounded: 1t2/3→∞\frac{1}{t^{2/3}} \to \infty and 1t1/3→∞\frac{1}{t^{1/3}} \to \infty.

Direct substitution gives an indeterminate form ∞∞\frac{\infty}{\infty}. To resolve this, eliminate the negative powers by multiplying the numerator and denominator by the dominant power with the largest magnitude, which is t2/3t^{2/3}:

g(t)=(5t−2/3+8t−1/3+12)⋅t2/3(3t−2/3−4t−1/3+2)⋅t2/3g(t) = \frac{\left(5 t^{-2/3} + 8 t^{-1/3} + 12\right) \cdot t^{2/3}}{\left(3 t^{-2/3} - 4 t^{-1/3} + 2\right) \cdot t^{2/3}}

Distribute t2/3t^{2/3} using the exponent addition rule ta⋅tb=ta+bt^a \cdot t^b = t^{a+b}:

  • t−2/3⋅t2/3=t0=1t^{-2/3} \cdot t^{2/3} = t^0 = 1
  • t−1/3⋅t2/3=t1/3t^{-1/3} \cdot t^{2/3} = t^{1/3}
  • 1⋅t2/3=t2/31 \cdot t^{2/3} = t^{2/3}

This rewrites g(t)g(t) as: g(t)=5+8t1/3+12t2/33−4t1/3+2t2/3g(t) = \frac{5 + 8t^{1/3} + 12t^{2/3}}{3 - 4t^{1/3} + 2t^{2/3}}

Now evaluate the limit as t→0+t \to 0^+. Since t1/3→0t^{1/3} \to 0 and t2/3→0t^{2/3} \to 0: lim⁡t→0+g(t)=5+8(0)+12(0)3−4(0)+2(0)=53\lim_{t \to 0^+} g(t) = \frac{5 + 8(0) + 12(0)}{3 - 4(0) + 2(0)} = \frac{5}{3}

Summary

Combining the two results into the requested ordered pair: (lim⁡t→∞g(t),lim⁡t→0+g(t))=(6,53)\left( \lim_{t \to \infty} g(t), \lim_{t \to 0^+} g(t) \right) = \left(6, \frac{5}{3}\right)

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student treats negative powers as positive powers as t → ∞, incorrectly taking the ratio of the coefficients of t^(-2/3) as the limit at infinity.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Partially correct
  1. Step 1

    As \text{As } t →\to ∞\infty,  the highest power terms dominate: \text{ the highest power terms dominate: } 5t−2/33t−2/3\frac{5t^{-2/3}}{3t^{-2/3}}   ⟹  \implies $\lim_{t →\to ∞\infty}$ g(t) = \frac{5}{3}_

    Correct

    $\lim_{t →\to ∞\infty} t^{-2/3} = 0 \text{ and } \\lim_{t \to \infty}t−1/3=0t^{-1/3} = 0\implies \\lim_{t →\to ∞\infty}$ g(t) = \frac{12}{2} = 6

    Because −2/3<0-2/3 < 0, t−2/3=1t2/3→0t^{-2/3} = \frac{1}{t^{2/3}} \to 0 as t→∞t \to \infty. Therefore, the variable terms vanish as t→∞t \to \infty, leaving the ratio of the constant terms 122=6\frac{12}{2} = 6.

  2. Step 2

    As \text{As } t →\to 0^+, t^{-2/3} →\to ∞\infty  and dominates\text{ and dominates}   ⟹  \implies $\lim_{t →\to 0^+}$ g(t) = \frac{5}{3}_

  3. Step 3

    (53,53)\left(\frac{5}{3}, \frac{5}{3}\right)

    Correct

    (6,53)\left(6, \frac{5}{3}\right)

    This ordered pair inherits the incorrect value from the limit as t→∞t \to \infty.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly reasoned that t−2/3→∞t^{-2/3} \to \infty as t→0+t \to 0^+ and determined the dominant behavior as t→0+t \to 0^+.

You correctly simplified the ratio of the leading coefficients to evaluate lim⁡t→0+g(t)=53\lim_{t \to 0^+} g(t) = \frac{5}{3}.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Negative Exponent Asymptotic Confusion

no

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

You applied the standard polynomial rule for t→∞t \to \infty without accounting for the fact that the exponents are negative. In a standard polynomial with positive powers, the term with the largest exponent dominates as t→∞t \to \infty; but for negative powers, t−p=1tp→0t^{-p} = \frac{1}{t^p} \to 0 as t→∞t \to \infty, meaning the constant term t0t^0 has the highest power and dominates.

REASONING PATH
Your ACEpath teacher’s suggested approach
Behavior of Negative Powers as $t \to \infty$

  • For any p>0p > 0, lim⁡t→∞t−p=lim⁡t→∞1tp=0\lim_{t \to \infty} t^{-p} = \lim_{t \to \infty} \frac{1}{t^p} = 0.

  • When evaluating limits as t→∞t \to \infty, set every term with a negative exponent directly to 0.

Identifying the Dominant Term

  • As t→∞t \to \infty, the term with the algebraically largest exponent dominates (e.g., 0>−1/3>−2/30 > -1/3 > -2/3).

  • Compare all exponents algebraically; constant terms correspond to t0t^0, which is larger than any negative power.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Rewrite negative powers as reciprocals: t−2/3=1t2/3t^{-2/3} = \frac{1}{t^{2/3}} and t−1/3=1t1/3t^{-1/3} = \frac{1}{t^{1/3}}.

  2. Evaluate the limit as t→∞t \to \infty: lim⁡t→∞g(t)=5(0)+8(0)+123(0)−4(0)+2=122=6\lim_{t \to \infty} g(t) = \frac{5(0) + 8(0) + 12}{3(0) - 4(0) + 2} = \frac{12}{2} = 6.

  3. Combine with your correct limit as t→0+t \to 0^+ to form the final ordered pair.

Final answer

(6,53)\left(6, \frac{5}{3}\right)

NEXT STEPS
Final Takeaways from Your Teacher

  • Always write negative powers as fractions 1tp\frac{1}{t^p} when checking behavior at infinity to avoid confusing them with positive powers.

  • Practice checking both limits t→∞t \to \infty and t→0+t \to 0^+ on rational expressions with mixed exponent signs.

OverviewACEpath Teacher's Diagnosis

02 The student confuses the behavior as t → 0^+ with t → ∞, incorrectly setting negative power terms to 0 as t → 0^+ to get 12/2 = 6.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Partially correct
  1. Step 1

    As t→∞t \to \infty, t−2/3→0t^{-2/3} \to 0 and t−1/3→0t^{-1/3} \to 0   ⟹  \implies lim⁡t→∞g(t)=122=6\lim_{t \to \infty} g(t) = \frac{12}{2} = 6

  2. Step 2

    As t→0+t \to 0^+, set t=0t=0 in each term with tt   ⟹  \implies lim⁡t→0+g(t)=122=6\lim_{t \to 0^+} g(t) = \frac{12}{2} = 6

    Correct

    Multiply numerator and denominator by t2/3t^{2/3} to clear negative powers: g(t)=5+8t1/3+12t2/33−4t1/3+2t2/3  ⟹  lim⁡t→0+g(t)=53g(t) = \frac{5 + 8t^{1/3} + 12t^{2/3}}{3 - 4t^{1/3} + 2t^{2/3}} \implies \lim_{t \to 0^+} g(t) = \frac{5}{3}

    Because t−p=1tpt^{-p} = \frac{1}{t^p}, as t→0+t \to 0^+, the terms t−2/3t^{-2/3} and t−1/3t^{-1/3} approach ∞\infty, not 00. Setting t=0t = 0 directly ignores the negative power and misses the dominant leading terms.

  3. Step 3

    (6, 6)

    Correct

    (6,53)\left(6, \frac{5}{3}\right)

    The second coordinate of your final ordered pair is incorrect due to the evaluated limit as t→0+t \to 0^+ in the previous step.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly evaluated the asymptotic limit as t→∞t \to \infty by using the property that t−p→0t^{-p} \to 0 for p>0p > 0.

You clearly organized your work and presented the final conclusion as an ordered pair.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Misapplying direct substitution to negative exponents near zero

no

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

You treated terms with negative rational exponents as if they had positive exponents when evaluating the limit as t→0+t \to 0^+. A negative exponent indicates a reciprocal (t−p=1tpt^{-p} = \frac{1}{t^p}), so as t→0+t \to 0^+, the term approaches ∞\infty rather than 00.

REASONING PATH
Your ACEpath teacher’s suggested approach
Check the sign of the exponent

  • For p>0p > 0, tp→0t^p \to 0 as t→0+t \to 0^+, but t−p=1tp→+∞t^{-p} = \frac{1}{t^p} \to +\infty.

  • Before substituting 00, rewrite every negative exponent as a fraction with a positive exponent in the denominator.

Clear negative exponents in rational expressions

  • When examining limits near 00 for expressions with negative powers, multiply numerator and denominator by tkt^k, where kk is the largest negative exponent magnitude.

  • Multiply the entire fraction by t2/3/t2/3t^{2/3}/t^{2/3} to obtain standard non-negative powers, then evaluate the limit.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Rewrite g(t)g(t) by multiplying the numerator and denominator by t2/3t^{2/3}: g(t)=(5t−2/3+8t−1/3+12)t2/3(3t−2/3−4t−1/3+2)t2/3=5+8t1/3+12t2/33−4t1/3+2t2/3g(t) = \frac{(5t^{-2/3} + 8t^{-1/3} + 12)t^{2/3}}{(3t^{-2/3} - 4t^{-1/3} + 2)t^{2/3}} = \frac{5 + 8t^{1/3} + 12t^{2/3}}{3 - 4t^{1/3} + 2t^{2/3}}

  2. Evaluate the limit as t→0+t \to 0^+ using the rewritten form: lim⁡t→0+5+8(0)1/3+12(0)2/33−4(0)1/3+2(0)2/3=53\lim_{t \to 0^+} \frac{5 + 8(0)^{1/3} + 12(0)^{2/3}}{3 - 4(0)^{1/3} + 2(0)^{2/3}} = \frac{5}{3}

  3. Combine the two limits into the requested ordered pair: (6,53)\left(6, \frac{5}{3}\right)

Final answer

(6,53)\left(6, \frac{5}{3}\right)

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice rewriting expressions with negative powers into fraction form before taking limits.

  • Review the dual behaviors of tpt^p and t−pt^{-p} under both t→∞t \to \infty and t→0+t \to 0^+.

OverviewACEpath Teacher's Diagnosis

03 The student swaps the roles of the limits at infinity and at zero, reversing the coordinates in the ordered pair.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Partially correct
  1. Step 1

    As t→∞t \to \infty, t−2/3→0t^{-2/3} \to 0 and t−1/3→0  ⟹  lim⁡t→∞g(t)=122=6t^{-1/3} \to 0 \implies \lim_{t \to \infty} g(t) = \frac{12}{2} = 6

  2. Step 2

    As t→0+t \to 0^+, multiply by t2/3t^{2/3} to get 5+8t1/3+12t2/33−4t1/3+2t2/3  ⟹  lim⁡t→0+g(t)=53\frac{5 + 8t^{1/3} + 12t^{2/3}}{3 - 4t^{1/3} + 2t^{2/3}} \implies \lim_{t \to 0^+} g(t) = \frac{5}{3}

  3. Step 3

    Ordered pair: (53,6)\text{Ordered pair: } \left(\frac{5}{3}, 6\right)

    Correct

    (6,53)\left(6, \frac{5}{3}\right)

    The prompt specifically asked for (lim⁡t→∞g(t),lim⁡t→0+g(t))\left(\lim_{t \to \infty} g(t), \lim_{t \to 0^+} g(t)\right), but you placed the limit as t→0+t \to 0^+ first and the limit as t→∞t \to \infty second.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly determined that negative powers tend to zero as t→∞t \to \infty.

You multiplied numerator and denominator by the dominant power t2/3t^{2/3} to resolve the indeterminate form as t→0+t \to 0^+.

You computed the exact numerical values for both individual limits without arithmetic errors.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Reversed Coordinate Order

You accidentally reversed the positions of the two limits when writing the requested ordered pair.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Overlooking the prompt's specified coordinate convention (lim⁡t→∞g(t),lim⁡t→0+g(t))\left(\lim_{t \to \infty} g(t), \lim_{t \to 0^+} g(t)\right) after completing the two separate calculations.

REASONING PATH
Your ACEpath teacher’s suggested approach
Coordinate Matching Check

  • An ordered pair (a,b)(a, b) is order-sensitive; the first coordinate must strictly match the first component requested by the prompt.

  • Before boxing your final answer, write down the template (t→∞,t→0+)(t \to \infty, t \to 0^+) and substitute the values directly into their assigned slots.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Identify the prompt's required output format: (lim⁡t→∞g(t),lim⁡t→0+g(t))\left(\lim_{t \to \infty} g(t), \lim_{t \to 0^+} g(t)\right).

  2. Substitute the value for the first component: lim⁡t→∞g(t)=6\lim_{t \to \infty} g(t) = 6.

  3. Substitute the value for the second component: lim⁡t→0+g(t)=53\lim_{t \to 0^+} g(t) = \frac{5}{3}.

Final answer

(6,53)\left(6, \frac{5}{3}\right)

NEXT STEPS
Final Takeaways from Your Teacher

  • Underline the target output format in the question stem before starting algebraic calculations.

  • Practice cross-referencing your final ordered pair coordinates against the exact order specified in multi-limit questions.

Lock it in

Question

What is the limit of t−pt^{-p} as t→∞t \to \infty when p>0p > 0?

Answer
  1. Since t−p=1tpt^{-p} = \frac{1}{t^p}, the denominator grows infinitely large while the numerator remains 1.

Similar question

01

Evaluate lim⁡t→∞4t−3/4+102t−3/4+5\lim_{t \to \infty} \frac{4t^{-3/4} + 10}{2t^{-3/4} + 5}.

Answer
  1. Key step: Since −3/4<0-3/4 < 0, lim⁡t→∞t−3/4=0\lim_{t \to \infty} t^{-3/4} = 0. Substituting gives 4(0)+102(0)+5=105=2\frac{4(0) + 10}{2(0) + 5} = \frac{10}{5} = 2.
02

Evaluate the asymptotic limits of h(x)=6x−1/2+32x−1/2+4h(x) = \frac{6x^{-1/2} + 3}{2x^{-1/2} + 4} as: (a) x→∞x \to \infty, and (b) x→0+x \to 0^+. Express your answer as (a,b)(a, b).

Answer

(34,3)\left(\frac{3}{4}, 3\right). Key steps: As x→∞x \to \infty, x−1/2→0x^{-1/2} \to 0, so lim⁡x→∞h(x)=34\lim_{x \to \infty} h(x) = \frac{3}{4}. As x→0+x \to 0^+, x−1/2→∞x^{-1/2} \to \infty dominates, so lim⁡x→0+h(x)=62=3\lim_{x \to 0^+} h(x) = \frac{6}{2} = 3.

03

Evaluate lim⁡t→∞3t1/2+7t−1/2+26t1/2−t−1/2+5\lim_{t \to \infty} \frac{3t^{1/2} + 7t^{-1/2} + 2}{6t^{1/2} - t^{-1/2} + 5} and lim⁡t→0+3t1/2+7t−1/2+26t1/2−t−1/2+5\lim_{t \to 0^+} \frac{3t^{1/2} + 7t^{-1/2} + 2}{6t^{1/2} - t^{-1/2} + 5}.

Answer

lim⁡t→∞=12\lim_{t \to \infty} = \frac{1}{2} and lim⁡t→0+=−7\lim_{t \to 0^+} = -7. Key steps: As t→∞t \to \infty, the highest power is t1/2t^{1/2}, so the ratio of leading coefficients gives 36=12\frac{3}{6} = \frac{1}{2}. As t→0+t \to 0^+, the most negative power t−1/2=1t1/2t^{-1/2} = \frac{1}{t^{1/2}} dominates, yielding 7−1=−7\frac{7}{-1} = -7.

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