If `alpha in(0,1)` and `f:R->R` and `lim_(x->oo)f(x)=0,lim_(x->oo)(f(x)-f(alphax))/x=0,` then `lim_(x->oo)f(x)/x=lambda` where `2lambda+7` is

Explanation

To solve the problem step by step, we will analyze the given limits and derive the required value of \(2\lambda + 7\).

### Step 1: Understand the Given Information
We have the following limits:
1. \(\lim_{x \to \infty} f(x) = 0\)
2. \(\lim_{x \to \infty} \frac{f(x) - f(\alpha x)}{x} = 0\)
3. We need to find \(\lim_{x \to \infty} \frac{f(x)}{x} = \lambda\)

### Step 2: Analyze the Second Limit
From the second limit, we can rewrite it as:
\[ \lim_{x \to \infty} \left( \frac{f(x)}{x} - \frac{f(\alpha x)}{x} \right) = 0 \]
This implies:
\[ \lim_{x \to \infty} \frac{f(x)}{x} = \lim_{x \to \infty} \frac{f(\alpha x)}{x} \]

### Step 3: Change of Variable
Let \(y = \alpha x\). As \(x \to \infty\), \(y \to \infty\) as well. Thus, we can express the second limit in terms of \(y\):
\[ \lim_{y \to \infty} \frac{f\left(\frac{y}{\alpha}\right)}{\frac{y}{\alpha}} = \alpha \lim_{y \to \infty} \frac{f\left(\frac{y}{\alpha}\right)}{y} \]

### Step 4: Substitute the Limits
Using the fact that \(\lim_{x \to \infty} \frac{f(x)}{x} = \lambda\), we have:
\[ \lambda = \alpha \lambda \]

### Step 5: Solve for \(\lambda\)
Rearranging gives:
\[ \lambda - \alpha \lambda = 0 \implies \lambda(1 - \alpha) = 0 \]
Since \(\alpha \in (0, 1)\), we must have \(\lambda = 0\).

### Step 6: Calculate \(2\lambda + 7\)
Now that we have \(\lambda = 0\), we can substitute it into the expression:
\[ 2\lambda + 7 = 2(0) + 7 = 7 \]

### Final Answer
Thus, the value of \(2\lambda + 7\) is:
\[ \boxed{7} \]

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