If `lim_(x to 1) sec^(-1)[lambda^(2)/(log x) - lambda^(2)/(x-1)]` exists then `lambda in`

Options

A. `[-infty, -sqrt(2)]`
B. `[sqrt(2), infty]`
C. `[-infty, - sqrt(2)] cup [sqrt(2), infty]`
D. none

Answer

To solve the limit problem, we need to analyze the expression given in the limit and find the conditions under which it exists. The limit we need to evaluate is:

\[ L = \lim_{x \to 1} \sec^{-1}\left[\frac{\lambda^2}{\log x} - \frac{\lambda^2}{x-1}\right] \]

### Step 1: Change of Variable
Let \( t = x - 1 \). As \( x \to 1 \), \( t \to 0 \). Therefore, we can rewrite the limit in terms of \( t \):

\[ L = \lim_{t \to 0} \sec^{-1}\left[\frac{\lambda^2}{\log(1+t)} - \frac{\lambda^2}{t}\right] \]

### Step 2: Expand \(\log(1+t)\)
Using the Taylor series expansion for \(\log(1+t)\) around \( t = 0 \):

\[ \log(1+t) \approx t - \frac{t^2}{2} + O(t^3) \]

Substituting this into our limit gives:

\[ L = \lim_{t \to 0} \sec^{-1}\left[\frac{\lambda^2}{t - \frac{t^2}{2} + O(t^3)} - \frac{\lambda^2}{t}\right] \]

### Step 3: Simplify the Expression
Now, simplifying the expression inside the secant inverse:

\[ \frac{\lambda^2}{t - \frac{t^2}{2}} - \frac{\lambda^2}{t} = \frac{\lambda^2 t - \lambda^2(t - \frac{t^2}{2})}{t(t - \frac{t^2}{2})} = \frac{\lambda^2 \frac{t^2}{2}}{t(t - \frac{t^2}{2})} \]

This simplifies to:

\[ \frac{\lambda^2}{2(t - \frac{t^2}{2})} \]

### Step 4: Evaluate the Limit
Now we can rewrite the limit as:

\[ L = \lim_{t \to 0} \sec^{-1}\left[\frac{\lambda^2}{2t}\right] \]

As \( t \to 0 \), \( \frac{\lambda^2}{2t} \to \infty \), and thus \( \sec^{-1}\left[\frac{\lambda^2}{2t}\right] \to \frac{\pi}{2} \) if \( \lambda^2 > 0 \).

### Step 5: Condition for Existence
For the limit to exist, the argument of secant inverse must be in the range of secant inverse function. The secant inverse function is defined for:

\[ x \leq -1 \quad \text{or} \quad x \geq 1 \]

Thus, we need:

\[ \frac{\lambda^2}{2t} \geq 1 \quad \text{as } t \to 0 \]

This implies:

\[ \lambda^2 \geq 2 \quad \Rightarrow \quad \lambda \geq \sqrt{2} \quad \text{or} \quad \lambda \leq -\sqrt{2} \]

### Final Result
Thus, the values of \( \lambda \) for which the limit exists are:

\[ \lambda \in (-\infty, -\sqrt{2}] \cup [\sqrt{2}, \infty) \]