Question:

Let f(x) and g(x) be differentiable functions on an open interval containing x = a, except possibly at x=a. Suppose that f(a) = g(a) = 0, g'(a) ≠ 0, and that the limit of f(x)/g(x) as x approaches a exists, but is of the form 0/0 or ∞/∞.

a) Use L'Hôpital's Rule to find an expression for the limit of f(x)/g(x) as x approaches a.

b) Evaluate the limit using L'Hôpital's Rule for the following function:

f(x) = x^2 - 3x + 2 g(x) = sin(x) - x

c) Determine whether the limit of f(x)/g(x) as x approaches a for the given function:

f(x) = (x - a)^3 g(x) = e^(3x) - e^(3a)

Answer:

a) L'Hôpital's Rule states that if the limit of f(x)/g(x) as x approaches a is of the form 0/0 or ∞/∞, and the functions f(x) and g(x) are differentiable on an open interval containing x = a, except possibly at x=a, and f(a) = g(a) = 0, g'(a) ≠ 0, then we can differentiate both f(x) and g(x) with respect to x and take the limit again:

lim(x→a) f(x)/g(x) = lim(x→a) f'(x)/g'(x)

b) We have the functions:

f(x) = x^2 - 3x + 2 g(x) = sin(x) - x

First, let's calculate the derivatives:

f'(x) = 2x - 3 g'(x) = cos(x) - 1

Now, applying L'Hôpital's Rule:

lim(x→a) f(x)/g(x) = lim(x→a) (2x - 3)/(cos(x) - 1)

To evaluate this limit, we substitute x = a:

lim(x→a) (2x - 3)/(cos(x) - 1) = (2a - 3)/(cos(a) - 1)

c) We have the functions:

f(x) = (x - a)^3 g(x) = e^(3x) - e^(3a)

First, let's calculate the derivatives:

f'(x) = 3(x - a)^2 g'(x) = 3e^(3x)

Now, applying L'Hôpital's Rule:

lim(x→a) f(x)/g(x) = lim(x→a) 3(x - a)^2/(3e^(3x))

To evaluate this limit, we substitute x = a:

lim(x→a) 3(x - a)^2/(3e^(3x)) = 3(a - a)^2/(3e^(3a))

Since (a - a)^2 = 0, we have:

lim(x→a) f(x)/g(x) = 0