AP Calculus AB Exam Question:
Use L'Hôpital's Rule to find the limit of the function as x approaches infinity:
x→∞lim3ex+x22ex+x3Step-by-step Solution:
To apply L'Hôpital's Rule, let's first take the derivatives of the numerator and denominator:
f(x)=2ex+x3g(x)=3ex+x2Taking the derivatives:
f′(x)=2ex+3x2g′(x)=3ex+2xNow, let's calculate the limit of the ratio of the derivatives:
x→∞limg′(x)f′(x)=x→∞lim3ex+2x2ex+3x2Since we still have an indeterminate form of ∞∞, we can once again apply L'Hôpital's Rule. Taking the derivatives again:
f′′(x)=2ex+6xg′′(x)=3ex+2Now, let's calculate the limit of the ratio of these new derivatives:
x→∞limg′′(x)f′′(x)=x→∞lim3ex+22ex+6xSince we don't have an indeterminate form anymore, we can directly evaluate the limit as x approaches infinity:
∞+2∞+0=∞∞=1Therefore, [ \lim_{{x \to \infty}} \frac{{2e^x + x^3}}{{3e^x + x^2}} = 1 ]
Answer: The limit of the given function as x approaches infinity is equal to 1.