Find the sum of infinite geometric series whose first term is f(x)=(1/1-x^9-7/1-x^7) as x tends to 1 and the common ratio is the limit of function g(x)=tanx-sinx/((ln(1+x))^3) as x-->0 is

To find the sum of the infinite geometric series defined by the first term \( f(x) \) and the common ratio derived from the limit of \( g(x) \), we need to break this down into manageable parts. Let's start with the first term and then move on to the common ratio.

Evaluating the First Term

The first term of the series is given by:

f(x) = (1/(1 - x^9)) - (7/(1 - x^7))

To find the sum as \( x \) approaches 1, we need to evaluate \( f(1) \). However, substituting \( x = 1 \) directly leads to an indeterminate form. Instead, we can use limits:

As \( x \) approaches 1:

• The term \( 1/(1 - x^9) \) approaches \( 1/(1 - 1) \), which is undefined.
• The term \( 7/(1 - x^7) \) also approaches \( 7/(1 - 1) \), which is similarly undefined.

To resolve this, we can apply L'Hôpital's Rule, which is useful for indeterminate forms of type \( \frac{0}{0} \). We differentiate the numerator and denominator separately:

For \( 1/(1 - x^9) \):

Let \( h(x) = 1 - x^9 \). The derivative \( h'(x) = -9x^8 \). Thus, applying L'Hôpital's Rule gives:

lim (x → 1) 1/h(x) = lim (x → 1) -1/(9x^8) = -1/9.

For \( 7/(1 - x^7) \):

Let \( k(x) = 1 - x^7 \). The derivative \( k'(x) = -7x^6 \). Applying L'Hôpital's Rule gives:

lim (x → 1) 7/k(x) = lim (x → 1) -7/(7x^6) = -1.

Now we can combine these results:

f(1) = -1/9 - (-1) = -1/9 + 1 = 8/9.

Finding the Common Ratio

Next, we need to determine the common ratio, which is the limit of the function:

g(x) = (tan x - sin x) / (ln(1 + x))^3 as x approaches 0.

Both the numerator and denominator approach 0 as \( x \) approaches 0, so we can again apply L'Hôpital's Rule.

First, we differentiate the numerator:

For \( tan x - sin x \), the derivative is \( sec^2 x - cos x \). Evaluating at \( x = 0 \) gives \( 1 - 1 = 0 \).

Now for the denominator, we differentiate \( (ln(1 + x))^3 \) using the chain rule:

Let \( u = ln(1 + x) \). Then the derivative is \( 3u^2 \cdot (1/(1 + x)) \). Evaluating at \( x = 0 \) gives \( 3(0)^2(1) = 0 \).

We apply L'Hôpital's Rule again:

For the numerator, differentiating \( sec^2 x - cos x \) gives \( 2sec^2 x \tan x + sin x \). Evaluating at \( x = 0 \) gives \( 2(1)(0) + 0 = 0 \).

For the denominator, differentiating \( 3u^2 \cdot (1/(1 + x)) \) gives \( 6u(1/(1 + x)) \cdot (1/(1 + x)) - 3u^2/(1 + x)^2 \). Evaluating at \( x = 0 \) gives \( 0 \).

Continuing this process, we find that the limit approaches 0, leading us to conclude that the common ratio is 0.

Calculating the Sum of the Series

Now that we have both components:

• First term \( a = 8/9 \)
• Common ratio \( r = 0 \)

The sum \( S \) of an infinite geometric series can be calculated using the formula:

S = a / (1 - r)

Substituting our values gives:

S = (8/9) / (1 - 0) = 8/9.

Thus, the sum of the infinite geometric series is \( \frac{8}{9} \).