RaySix

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Evaluating Limit as x Approaches Infinity

Created by @nathanedwards

at November 3rd 2023, 1:25:58 am.

AP_Calculus_AB-Questions

#calculus

#limits

#asymptote

AP Calculus AB Exam Question:

Evaluate the following limit as x approaches positive infinity:

\lim_{{x \to \infty}} \left(\frac{3x^2 + 5x}{2x^3 - 7}\right)

Step-by-step Solution:

To evaluate this limit, we can divide both the numerator and denominator by the highest exponent of x, which is x^3. This helps us to simplify the expression.

Dividing each term in the numerator and denominator by x^3, we get:

\lim_{{x \to \infty}} \left(\frac{3x^2/x^3 + 5x/x^3}{2x^3/x^3 - 7/x^3}\right)

Simplifying further, we have:

\lim_{{x \to \infty}} \left(\frac{3/x + 5/x^2}{2 - 7/x^3}\right)

Now, as x approaches positive infinity, 1/x approaches 0. Therefore, we can substitute 0 for 1/x in the expression:

\lim_{{x \to \infty}} \left(\frac{3(0) + 5(0^2)}{2 - 7(0^3)}\right)

Simplifying, we have:

\lim_{{x \to \infty}} \left(\frac{0}{2 - 0}\right)

\lim_{{x \to \infty}} \left(\frac{0}{2}\right)

Since the numerator of the fraction is 0, the value of the entire expression is 0 when x approaches positive infinity.

Therefore, the value of the limit is 0.

$\therefore \lim_{{x \to \infty}} \left(\frac{3x^2 + 5x}{2x^3 - 7}\right) = 0$