If `f(x) = lim_(n->oo) sum_(r=0)^n (tan(x/2^(r+1)) + tan^3 (x/2^(r+1)))/(1- tan^2 (x/2^(r+1)))` then `lim_(x->0) f(x)/x` is

Explanation

To solve the problem, we need to evaluate the limit given in the question step by step.

### Step 1: Define the function
We start with the function defined as:

\[ f(x) = \lim_{n \to \infty} \sum_{r=0}^{n} \frac{\tan\left(\frac{x}{2^{r+1}}\right) + \tan^3\left(\frac{x}{2^{r+1}}\right)}{1 - \tan^2\left(\frac{x}{2^{r+1}}\right)} \]

### Step 2: Simplify the expression
Using the identity for tangent, we can rewrite the expression inside the limit. We know that:

\[ \frac{\tan A + \tan^3 A}{1 - \tan^2 A} = \tan A \cdot \frac{1 + \tan^2 A}{1 - \tan^2 A} \]

Thus, we can express \(f(x)\) as:

\[ f(x) = \lim_{n \to \infty} \sum_{r=0}^{n} \tan\left(\frac{x}{2^{r+1}}\right) \cdot \frac{1 + \tan^2\left(\frac{x}{2^{r+1}}\right)}{1 - \tan^2\left(\frac{x}{2^{r+1}}\right)} \]

### Step 3: Change of variable
Let \( \alpha_r = \frac{x}{2^{r+1}} \). As \( r \) increases, \( \alpha_r \) approaches 0. Thus, we can rewrite the limit:

\[ f(x) = \lim_{n \to \infty} \sum_{r=0}^{n} \tan(\alpha_r) \cdot \frac{1 + \tan^2(\alpha_r)}{1 - \tan^2(\alpha_r)} \]

### Step 4: Use Taylor series expansion
For small values of \( \alpha_r \), we can use the Taylor series expansion of \(\tan\):

\[ \tan(\alpha_r) \approx \alpha_r + \frac{\alpha_r^3}{3} + O(\alpha_r^5) \]

### Step 5: Evaluate the limit
Substituting \(\alpha_r\) back into the function, we have:

\[ f(x) = \lim_{n \to \infty} \sum_{r=0}^{n} \left( \frac{x}{2^{r+1}} + \frac{1}{3}\left(\frac{x}{2^{r+1}}\right)^3 \right) \]

This sum can be evaluated using the formula for the sum of a geometric series. The first term converges to:

\[ \sum_{r=0}^{n} \frac{x}{2^{r+1}} = x \sum_{r=0}^{n} \frac{1}{2^{r+1}} = x \left(1 - \frac{1}{2^{n+1}}\right) \frac{1}{1 - \frac{1}{2}} = x \left(1 - \frac{1}{2^{n+1}}\right) \cdot 2 \]

As \( n \to \infty \), this approaches \( x \cdot 2 \).

### Step 6: Final limit evaluation
Thus, we find that:

\[ f(x) \approx 2x \]

Now we need to evaluate:

\[ \lim_{x \to 0} \frac{f(x)}{x} = \lim_{x \to 0} \frac{2x}{x} = 2 \]

### Conclusion
The final result is:

\[ \lim_{x \to 0} \frac{f(x)}{x} = 2 \]