AP Calculus AB Exam Question:

Consider the function f(x) = (1 - e^(2x))/(x^2 - 4x + 3).

(a) Determine if the limit of f(x) as x approaches 2 exists or not. Justify your answer.

(b) If the limit in part (a) exists, find its value using L'Hôpital's Rule.

Answer:

(a) To determine if the limit of f(x) as x approaches 2 exists, we evaluate the left-hand limit and right-hand limit separately:

Left-hand limit: lim(x -> 2-) (1 - e^(2x))/(x^2 - 4x + 3)

Substituting x = 2 - h: lim(h -> 0-) (1 - e^(2(2-h)))/((2-h)^2 - 4(2-h) + 3) = lim(h -> 0-) (1 - e^(4-2h))/(4 - 4h + h^2 - 8 + 4h + 3) = lim(h -> 0-) (1 - e^(4-2h))/(h^2 - 1)

Notice that in the numerator, as h approaches 0, the exponent, 4 - 2h, tends towards 4. Thus, the numerator approaches 1 - e^4.

In the denominator, as h approaches 0, the expression h^2 - 1 approaches -1.

Therefore, the left-hand limit is: lim(x -> 2-) (1 - e^(2x))/(x^2 - 4x + 3) = (1 - e^4)/(-1) = e^4 - 1.

Right-hand limit: lim(x -> 2+) (1 - e^(2x))/(x^2 - 4x + 3)

Substituting x = 2 + h: lim(h -> 0+) (1 - e^(2(2+h)))/((2+h)^2 - 4(2+h) + 3) = lim(h -> 0+) (1 - e^(4+2h))/(4 + 4h + h^2 - 8 - 4h + 3) = lim(h -> 0+) (1 - e^(4+2h))/(h^2 - 1)

Similar to the left-hand limit, in the numerator 1 - e^(4+2h) approaches 1 - e^4 as h approaches 0.

In the denominator, h^2 - 1 still approaches -1 as h approaches 0.

Therefore, the right-hand limit is: lim(x -> 2+) (1 - e^(2x))/(x^2 - 4x + 3) = (1 - e^4)/(-1) = e^4 - 1.

Now, comparing the left-hand limit and the right-hand limit, we find that both limits are equal to e^4 - 1.

Since the left-hand limit and right-hand limit are equal, the overall limit of f(x) as x approaches 2 exists and is equal to e^4 - 1.

(b) Using L'Hôpital's Rule:

To find the limit of f(x) as x approaches 2, we differentiate the numerator and denominator separately and then evaluate the limit again.

Differentiating the numerator: f'(x) = (0 - 2e^(2x))/(x^2 - 4x + 3)

Differentiating the denominator: g'(x) = 2x - 4

Now, we can evaluate the limit of f'(x)/g'(x) as x approaches 2: lim(x -> 2) [(0 - 2e^(2x))/(x^2 - 4x + 3)] / (2x - 4) = lim(x -> 2) (-2e^(4))/(4 - 4) = -1/2 * e^4

Therefore, using L'Hôpital's Rule, the limit of f(x) as x approaches 2 is -1/2 * e^4.

In conclusion, the limit of f(x) as x approaches 2 exists and is equal to -1/2 * e^4.