Q 63. In the case where n' is a positive integer and limx—> 7(x^n-7^n)/(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), and the derivative of 7^n is 0 (since it's a constant). So, the derivative of the numerator (x^n - 7^n) is nx^(n-1) - 0 = n*x^(n-1).

The derivative of x is 1, and the derivative of 7 is 0. So, the derivative of the denominator (x - 7) is 1 - 0 = 1.

Now, let's apply L'Hopital's Rule:

lim (x→7) [(x^n - 7^n) / (x - 7)] = lim (x→7) [(nx^(n-1)) / 1] = n7^(n-1).

We know from the problem that this limit is equal to 147. So, we have the equation:

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

Solving this equation will give us the value of n. Let's try the options:

A. n=3: 37^(3-1) = 349 = 147. This is correct.

B. n=1: 17^(1-1) = 11 = 1. This is not correct.

C. n=2: 27^(2-1) = 27 = 14. This is not correct.

D. n=4: 47^(4-1) = 4343 = 1372. This is not correct.

So, the correct answer is A. n=3.