if `|x| <1 ` then `(L t)_(n->oo)(1+x)(1+x^2)(1+x^4)........(1+x^(2n))=`

Options

A. `1/(x-1)`
B. `1/(1-x)`
C. `1-x`
D. `x-1`

Answer

To solve the problem, we need to evaluate the limit:

\[ \lim_{n \to \infty} (1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^n}) \]

given that \(|x| < 1\).

### Step 1: Rewrite the expression

We start by rewriting the product:

\[ P_n = (1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^n}) \]

### Step 2: Multiply and divide by \(1-x\)

Next, we multiply \(P_n\) by \(1-x\) and then divide by \(1-x\):

\[ (1-x)P_n = (1-x)(1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^n}) \]

### Step 3: Simplify the expression

Now we can simplify the left-hand side:

\[ (1-x)(1+x) = 1 - x^2 \]

Continuing this process, we can see that:

\[ (1-x)(1+x^2) = 1 - x^4 \]

Following this pattern, we can express the entire product as:

\[ (1-x)(1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^n}) = 1 - x^{2^{n+1}} \]

### Step 4: Write the final expression

Thus, we have:

\[ (1-x)P_n = 1 - x^{2^{n+1}} \]

### Step 5: Solve for \(P_n\)

Now, we can solve for \(P_n\):

\[ P_n = \frac{1 - x^{2^{n+1}}}{1 - x} \]

### Step 6: Take the limit as \(n \to \infty\)

Now we need to evaluate the limit as \(n\) approaches infinity:

\[ \lim_{n \to \infty} P_n = \lim_{n \to \infty} \frac{1 - x^{2^{n+1}}}{1 - x} \]

Since \(|x| < 1\), \(x^{2^{n+1}} \to 0\) as \(n \to \infty\). Therefore, we have:

\[ \lim_{n \to \infty} P_n = \frac{1 - 0}{1 - x} = \frac{1}{1 - x} \]

### Final Answer

Thus, the limit is:

\[ \lim_{n \to \infty} (1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^n}) = \frac{1}{1 - x} \]

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