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

Options

A. 1
B. 0
C. -1
D. none of these

Explanation

To solve the problem step by step, we will analyze the given function and compute the limit as required.

### Step 1: Define the function
We start with the function:
\[ f(x) = \lim_{n \to \infty} \sum_{r=0}^{n} \frac{\tan\left(\frac{x}{2^{r+1}}\right) + \tan^3\left(\frac{x}{2^{r+1}}\right)}{1 - \tan^2\left(\frac{x}{2^{r+1}}\right)} \]

### Step 2: Simplify the expression inside the limit
We can rewrite the expression using the identity for tangent:
\[ \tan(a) + \tan^3(a) = \tan(a)(1 + \tan^2(a)) = \tan(a) \sec^2(a) \]
Thus, we have:
\[ f(x) = \lim_{n \to \infty} \sum_{r=0}^{n} \tan\left(\frac{x}{2^{r+1}}\right) \sec^2\left(\frac{x}{2^{r+1}}\right) \]

### Step 3: Substitute \(\alpha_r\)
Let \(\alpha_r = \frac{x}{2^{r+1}}\). Then, as \(n \to \infty\), \(\alpha_r \to 0\) for each \(r\). Therefore, we can express \(f(x)\) as:
\[ f(x) = \lim_{n \to \infty} \sum_{r=0}^{n} \tan(\alpha_r) \sec^2(\alpha_r) \]

### Step 4: Apply Taylor expansion for small angles
For small angles, we have the approximations:
\[ \tan(\alpha_r) \approx \alpha_r \quad \text{and} \quad \sec^2(\alpha_r) \approx 1 \]
Thus:
\[ f(x) \approx \lim_{n \to \infty} \sum_{r=0}^{n} \alpha_r = \lim_{n \to \infty} \sum_{r=0}^{n} \frac{x}{2^{r+1}} = x \lim_{n \to \infty} \sum_{r=0}^{n} \frac{1}{2^{r+1}} \]

### Step 5: Evaluate the geometric series
The sum \(\sum_{r=0}^{n} \frac{1}{2^{r+1}}\) is a geometric series with first term \(\frac{1}{2}\) and common ratio \(\frac{1}{2}\):
\[ \sum_{r=0}^{\infty} \frac{1}{2^{r+1}} = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 1 \]
Thus:
\[ f(x) = x \cdot 1 = x \]

### Step 6: Find the limit
Now we need to find:
\[ \lim_{x \to 0} \frac{f(x)}{x} = \lim_{x \to 0} \frac{x}{x} = 1 \]

### Conclusion
Therefore, the final answer is:
\[ \lim_{x \to 0} \frac{f(x)}{x} = 1 \]

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