Q 63. In the case where n' is a positive integer and limx—7(x"-7")/(x-7) = 147, then n is equals Ops: A. n=3 B. n=1 C. n=2 D. n=4

This problem can be solved using L'Hopital's Rule, which states that the limit of a quotient of two functions as x approaches a certain value is equal to the limit of the quotients of their derivatives.

First, let's differentiate the numerator and the denominator:

The derivative of x^n is nx^(n-1), so the derivative of x^n - 7^n is nx^(n-1) - 0 (since 7^n is a constant and its derivative is 0).

The derivative of x - 7 is 1.

So, the limit becomes lim(x→7) [n*x^(n-1)] / 1 = 147.

Substituting x = 7, we get n*7^(n-1) = 147.

We can simplify this equation by dividing both sides by 7^(n-1):

n = 147 / 7^(n-1).

Now, we can try the options given to see which one satisfies this equation:

A. n = 3: 147 / 7^(3-1) = 147 / 49 = 3. This does not equal to 3. B. n = 1: 147 / 7^(1-1) = 147 / 1 = 147. This does not equal to 1. C. n = 2: 147 / 7^(2-1) = 147 / 7 = 21. This does not equal to 2. D. n = 4: 147 / 7^(4-1) = 147 / 343 = 0.428. This does not equal to 4.

None of the options satisfy the equation, so there seems to be a mistake in the problem or the options given.