Select the correct alternatives out of given four alternatives in each.
`lim_(xto(pi)/(4))(sec^(2)x-2)/(tanx-1)` is-

Options

A. 0
B. 1
C. 2
D. 4

Answer

To solve the limit \( \lim_{x \to \frac{\pi}{4}} \frac{\sec^2 x - 2}{\tan x - 1} \), we will follow these steps:

### Step 1: Evaluate the limit directly
First, we substitute \( x = \frac{\pi}{4} \) into the expression:

\[ \sec^2\left(\frac{\pi}{4}\right) = \frac{1}{\cos^2\left(\frac{\pi}{4}\right)} = \frac{1}{\left(\frac{1}{\sqrt{2}}\right)^2} = 2 \]

And,

\[ \tan\left(\frac{\pi}{4}\right) = 1 \]

Substituting these values into the limit gives:

\[ \frac{\sec^2\left(\frac{\pi}{4}\right) - 2}{\tan\left(\frac{\pi}{4}\right) - 1} = \frac{2 - 2}{1 - 1} = \frac{0}{0} \]

This is an indeterminate form.

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that if \( \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0} \), then:

\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]

Here, let \( f(x) = \sec^2 x - 2 \) and \( g(x) = \tan x - 1 \).

### Step 3: Differentiate the numerator and denominator
Now we differentiate \( f(x) \) and \( g(x) \):

1. The derivative of \( f(x) = \sec^2 x - 2 \):
\[ f'(x) = 2 \sec^2 x \tan x \]

2. The derivative of \( g(x) = \tan x - 1 \):
\[ g'(x) = \sec^2 x \]

### Step 4: Substitute back into the limit
Now we substitute these derivatives back into the limit:

\[ \lim_{x \to \frac{\pi}{4}} \frac{2 \sec^2 x \tan x}{\sec^2 x} \]

This simplifies to:

\[ \lim_{x \to \frac{\pi}{4}} 2 \tan x \]

### Step 5: Evaluate the limit again
Now we substitute \( x = \frac{\pi}{4} \):

\[ 2 \tan\left(\frac{\pi}{4}\right) = 2 \cdot 1 = 2 \]

### Conclusion
Thus, the limit is:

\[ \lim_{x \to \frac{\pi}{4}} \frac{\sec^2 x - 2}{\tan x - 1} = 2 \]

### Final Answer
The correct alternative is \( 2 \).

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