Post

Created by @nathanedwards
 at November 2nd 2023, 4:39:31 am.

AP Calculus AB Exam Question:

Use L'Hôpital's Rule to find the limit of the function as x approaches infinity:

limx2ex+x33ex+x2 \lim_{{x \to \infty}} \frac{{2e^x + x^3}}{{3e^x + x^2}}

Step-by-step Solution:

To apply L'Hôpital's Rule, let's first take the derivatives of the numerator and denominator:

f(x)=2ex+x3 f(x) = 2e^x + x^3
g(x)=3ex+x2 g(x) = 3e^x + x^2

Taking the derivatives:

f(x)=2ex+3x2 f'(x) = 2e^x + 3x^2
g(x)=3ex+2x g'(x) = 3e^x + 2x

Now, let's calculate the limit of the ratio of the derivatives:

limxf(x)g(x)=limx2ex+3x23ex+2x \lim_{{x \to \infty}} \frac{{f'(x)}}{{g'(x)}} = \lim_{{x \to \infty}} \frac{{2e^x + 3x^2}}{{3e^x + 2x}}

Since we still have an indeterminate form of \frac{\infty}{\infty}, we can once again apply L'Hôpital's Rule. Taking the derivatives again:

f(x)=2ex+6x f''(x) = 2e^x + 6x
g(x)=3ex+2 g''(x) = 3e^x + 2

Now, let's calculate the limit of the ratio of these new derivatives:

limxf(x)g(x)=limx2ex+6x3ex+2 \lim_{{x \to \infty}} \frac{{f''(x)}}{{g''(x)}} = \lim_{{x \to \infty}} \frac{{2e^x + 6x}}{{3e^x + 2}}

Since we don't have an indeterminate form anymore, we can directly evaluate the limit as x approaches infinity:

+0+2==1 \frac{{\infty + 0}}{{\infty + 2}} = \frac{\infty}{\infty} = 1

Therefore, [ \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.