In order to solve this problem, you can make use of L'Hopital's Rule.
L'Hopital's rule states that the limit as x approaches a certain value of the ratio of two functions, where both the numerator and denominator tend to zero or infinity, is equal to the limit of the ratio of their derivatives.
Here we wish to find:
lim (x→0) (1−cos(9x))/(1−cos(8x))
First, it is necessary to calculate the derivative of both the numerator and the denominator:
The derivative of the function 1 - cos(9x) is sin(9x) * 9.
The derivative of the function 1 - cos(8x) is sin(8x) * 8.
So, we now have to calculate the limit of the ratio of the new functions as x tends to 0.
So we now have lim (x→0) [(sin(9x) * 9) / (sin(8x) * 8)]
Using the property of limits where lim (x→a) [f(x) / g(x)] = [lim (x→a) f(x)] / [lim (x→a) g(x)], we get:
lim (x→0) sin(9x) * 9 / lim (x→0) sin(8x) * 8,
This simplifies to:
(9 * lim (x→0) sin(9x)) / (8 * lim (x→0) sin(8x)).
As x tends to 0, sin(ax) tends to 0 for all finite a, so we end up with:
(9 * 0) / (8 * 0) = 0/0.
0/0 is an indeterminate form, so we apply L'Hopital's rule once again and take the derivatives of the numerator and the denominator again:
The derivative of the function sin(9x) is cos(9x) * 9.
The derivative of the function sin(8x) is cos(8x) * 8.
So, we need to compute lim (x→0) [(cos(9x) * 9^2) / (cos(8x) * 8^2)].
As x approaches 0, cos(ax) approaches 1 for all finite a. Therefore, we can conclude that the limit is (9^2) / (8^2) = 81/64.