AP Calculus AB Exam Question

Consider the function f(x) defined by:

f(x) = (2x^3 + 5x^2 - 3x) / (4x^3 + 2x^2 - 5)

Determine the following limits as x approaches positive and negative infinity:

1. $\lim_{{x \to \infty}} f(x)$
2. $\lim_{{x \to -\infty}} f(x)$

Clearly explain each step of your solution.

Solution

1. $\lim_{{x \to \infty}} f(x)$:

To evaluate this limit, we need to examine the behavior of the function as x becomes very large.

Notice that the degree of the numerator (3) is the same as the degree of the denominator (3). To analyze this limit, we can focus on the terms with the highest degree in the numerator and denominator.

Let's rewrite the function by dividing both the numerator and denominator by x^3:

f(x) = (2x^3 + 5x^2 - 3x) / (4x^3 + 2x^2 - 5) = (2 + 5/x - 3/x^2) / (4 + 2/x - 5/x^3)

As x approaches positive infinity, the terms with 1/x^2 and 1/x^3 become negligible compared to the constant terms 2 and 4. This is because as x becomes larger, the denominator x^2 and x^3 grow much faster than the numerator terms.

Thus, the limit simplifies to:

lim_{{x \to \infty}} f(x) = (2 + 0 - 0) / (4 + 0 - 0) = 2/4 = 1/2

Therefore, $\lim_{{x \to \infty}} f(x) = \frac{1}{2}$.

2. $\lim_{{x \to -\infty}} f(x)$:

Similarly, to evaluate this limit, we need to examine the behavior of the function as x becomes very large in the negative direction.

Again, the degree of the numerator (3) is the same as the degree of the denominator (3). So, we focus on the highest degree terms.

Dividing both the numerator and denominator by x^3:

f(x) = (2x^3 + 5x^2 - 3x) / (4x^3 + 2x^2 - 5) = (2 + 5/x - 3/x^2) / (4 + 2/x - 5/x^3)

As x tends to negative infinity, the terms with 1/x^2 and 1/x^3 become negligible compared to the constant terms 2 and 4. This is because as x becomes very large in the negative direction, the denominator x^2 and x^3 grow much faster than the numerator terms.

Thus, the limit simplifies to:

lim_{{x \to -\infty}} f(x) = (2 + 0 - 0) / (4 + 0 - 0) = 2/4 = 1/2

Therefore, $\lim_{{x \to -\infty}} f(x) = \frac{1}{2}$.

In conclusion, both $\lim_{{x \to \infty}} f(x)$ and $\lim_{{x \to -\infty}} f(x)$ are equal to $\frac{1}{2}$.