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The limit of sqrt(9x^4 + 1)/(x^2 - 3x + 5) as x approaches infinity is


\displaystyle\lim_(x \to \infty) (√(9x^4+1))/(x^2 -3x + 5)


(A) 1

(B) 3

(C) 9

(D) nonexistent

User ThG
by
7.9k points

2 Answers

3 votes

Final answer:

The limit of
\(√(9x^4 + 1)/(x^2 - 3x + 5)\) as x approaches infinity is 3, after comparing the highest powers of x in both the numerator and the denominator and simplifying.

Step-by-step explanation:

To find the limit of the given function
\(\displaystyle\lim_(x \to \infty) (√(9x^4+1))/(x^2 -3x + 5)\) as x approaches infinity, we can use the property of limits involving infinity. We need to compare the highest powers of x in both the numerator and the denominator. The highest power of x in the numerator under the square root is x⁴, and outside the square root, it will be . In the denominator, the highest power is . If we divide the numerator and the denominator by , we get:


\[ (√(9x^4+1))/(x^2 -3x + 5) = \frac{\sqrt{(9x^4)/(x^4)+(1)/(x^4)}}{(x^2)/(x^2) -(3x)/(x^2) + (5)/(x^2)} = \frac{\sqrt{9+(1)/(x^4)}}{1 -(3)/(x) + (5)/(x^2)} \]

As x approaches infinity, the terms
\((1)/(x^4)\), \((3)/(x)\), and \((5)/(x^2)\) approach zero, and we are left with:


\[ (√(9))/(1) = 3 \]

Therefore, the limit of the given function as x approaches infinity is 3, which corresponds to option (B).

User Clayton Graul
by
7.7k points
3 votes

Answer:

B. 3.

Step-by-step explanation:

At the limit we can take the numerator to be √(9x^4) = 3x^2

The function is of the form ∞/ ∞ as x approaches ∞ so we can apply l'hopitals rule:

Differentiating top and bottom we have 6x / 2x - 3. Differentiating again we get 6 / 2 = 3.

Our limit as x approaches infinity is 3.

User ElderBug
by
7.8k points
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