Given , G(x)=-√25-x²therefore limit x==>1. G(x)-G(1)/x-1=limit x===>1 G`(x)-0/1-0using L Hospital's rule =G`(1)=1/√24 [Because G(x)=-✓25-x²==>G`(x)=2x/2√25-x²]

Given:
G(x) = -√(25-x²)

To find:
The value of the limit as x approaches 1 of [G(x)-G(1)] / (x-1)

Solution:

Step 1: Find G(1)
To find G(1), substitute x = 1 into the equation for G(x):

G(1) = -√(25-1²)
= -√(24)
= -√(4 * 6)
= -2√6

Step 2: Simplify the expression G(x)-G(1)
Substitute the values of G(x) and G(1) into the expression:

G(x) - G(1) = -√(25-x²) - (-2√6)
= -√(25-x²) + 2√6

Step 3: Simplify the expression (G(x)-G(1)) / (x-1)
Substitute the values of G(x) - G(1) and x - 1 into the expression:

[G(x) - G(1)] / (x - 1) = [-√(25-x²) + 2√6] / (x - 1)

Step 4: Apply the limit as x approaches 1
To find the value of the limit as x approaches 1, substitute x = 1 into the expression:

lim(x→1) [G(x) - G(1)] / (x - 1) = [-√(25-1²) + 2√6] / (1 - 1)

Since the denominator is 0, we can't directly substitute x = 1. However, we can simplify the expression further to determine the limit.

Step 5: Simplify the expression further
Simplify the expression by rationalizing the numerator:

lim(x→1) [G(x) - G(1)] / (x - 1) = [-√(25-x²) + 2√6] / (1 - 1)
= [-√(25-x²) + 2√6] / 0
= ∞

The limit as x approaches 1 of [G(x) - G(1)] / (x - 1) is equal to infinity.

Answer:
(D) none of these