The value of `lim_(xto(1)/(sqrt(2))) (x-cos(sin^(-1)x))/(1-tan(sin^(-1)x))" is "`

Options

A. `-(1)/(sqrt(2))`
B. -1
C. non-existent
D. none of these

Explanation

To solve the limit \( \lim_{x \to \frac{1}{\sqrt{2}}} \frac{x - \cos(\sin^{-1} x)}{1 - \tan(\sin^{-1} x)} \), we will follow these steps:

### Step 1: Substitute \( \sin^{-1} x \) with \( t \)
Let \( t = \sin^{-1} x \). Then, we have:
\[ x = \sin t \]
As \( x \to \frac{1}{\sqrt{2}} \), we find \( t \):
\[ t = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4} \]

### Step 2: Rewrite the limit in terms of \( t \)
Substituting \( x \) in the limit gives:
\[ \lim_{t \to \frac{\pi}{4}} \frac{\sin t - \cos t}{1 - \tan t} \]

### Step 3: Evaluate the limit
Now, we substitute \( t = \frac{\pi}{4} \):
\[ \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}, \quad \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}, \quad \tan\left(\frac{\pi}{4}\right) = 1 \]
Thus, we have:
\[ \frac{\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}}{1 - 1} = \frac{0}{0} \]
This is an indeterminate form, so we can apply L'Hôpital's Rule.

### Step 4: Apply L'Hôpital's Rule
We differentiate the numerator and denominator:
- The derivative of the numerator \( \sin t - \cos t \) is \( \cos t + \sin t \).
- The derivative of the denominator \( 1 - \tan t \) is \( -\sec^2 t \).

Thus, we rewrite the limit:
\[ \lim_{t \to \frac{\pi}{4}} \frac{\cos t + \sin t}{-\sec^2 t} \]

### Step 5: Substitute \( t = \frac{\pi}{4} \) again
Now substituting \( t = \frac{\pi}{4} \):
\[ \cos\left(\frac{\pi}{4}\right) + \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2} \]
And,
\[ -\sec^2\left(\frac{\pi}{4}\right) = -2 \]
Thus, the limit becomes:
\[ \frac{\sqrt{2}}{-2} = -\frac{\sqrt{2}}{2} \]

### Final Answer
The value of the limit is:
\[ \lim_{x \to \frac{1}{\sqrt{2}}} \frac{x - \cos(\sin^{-1} x)}{1 - \tan(\sin^{-1} x)} = -\frac{\sqrt{2}}{2} \]