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Math  /  Calculus

Questionlim⁡t→∞8t3+t(2t−1)(2t2+1)\lim _{t \rightarrow \infty} \frac{8 t^{3}+t}{(2 t-1)\left(2 t^{2}+1\right)}

Studdy Solution

STEP 1

1. We are dealing with a limit as t t approaches infinity.
2. The expression is a rational function, which is a ratio of two polynomials.
3. The behavior of the limit as t→∞ t \to \infty is determined by the highest degree terms in the numerator and denominator.

STEP 2

1. Identify the highest degree terms in the numerator and the denominator.
2. Simplify the expression by dividing both the numerator and the denominator by the highest power of t t present in the denominator.
3. Evaluate the limit of the simplified expression as t→∞ t \to \infty .

STEP 3

Identify the highest degree terms in the numerator and the denominator.
- The numerator is 8t3+t 8t^3 + t . The highest degree term is 8t3 8t^3 . - The denominator is (2t−1)(2t2+1) (2t-1)(2t^2+1) . First, expand this product:
(2t−1)(2t2+1)=2t⋅2t2+2t⋅1−1⋅2t2−1⋅1=4t3+2t−2t2−1(2t-1)(2t^2+1) = 2t \cdot 2t^2 + 2t \cdot 1 - 1 \cdot 2t^2 - 1 \cdot 1 = 4t^3 + 2t - 2t^2 - 1
The highest degree term in the expanded form is 4t3 4t^3 .

STEP 4

Simplify the expression by dividing both the numerator and the denominator by t3 t^3 , the highest power of t t in the denominator.
8t3+t4t3+2t−2t2−1=8t3t3+tt34t3t3+2tt3−2t2t3−1t3=8+1t24+2t2−2t−1t3\frac{8t^3 + t}{4t^3 + 2t - 2t^2 - 1} = \frac{\frac{8t^3}{t^3} + \frac{t}{t^3}}{\frac{4t^3}{t^3} + \frac{2t}{t^3} - \frac{2t^2}{t^3} - \frac{1}{t^3}} = \frac{8 + \frac{1}{t^2}}{4 + \frac{2}{t^2} - \frac{2}{t} - \frac{1}{t^3}}

STEP 5

Evaluate the limit of the simplified expression as t→∞ t \to \infty .
As t→∞ t \to \infty , the terms 1t2\frac{1}{t^2}, 2t2\frac{2}{t^2}, 2t\frac{2}{t}, and 1t3\frac{1}{t^3} all approach 0. Therefore, the expression simplifies to:
lim⁡t→∞8+1t24+2t2−2t−1t3=8+04+0−0−0=84=2\lim_{t \to \infty} \frac{8 + \frac{1}{t^2}}{4 + \frac{2}{t^2} - \frac{2}{t} - \frac{1}{t^3}} = \frac{8 + 0}{4 + 0 - 0 - 0} = \frac{8}{4} = 2
The limit is 2 \boxed{2} .

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