Using L' Hospital' s rule prove that `lim_(x to (pi)/(2))` (sec x - tan x ) = 0

Answer

To prove that \(\lim_{x \to \frac{\pi}{2}} (\sec x - \tan x) = 0\) using L'Hôpital's rule, we can follow these steps:

### Step 1: Rewrite the expression
We start with the limit:
\[ \lim_{x \to \frac{\pi}{2}} (\sec x - \tan x) \]
We know that \(\sec x = \frac{1}{\cos x}\) and \(\tan x = \frac{\sin x}{\cos x}\). Therefore, we can rewrite the expression as:
\[ \sec x - \tan x = \frac{1}{\cos x} - \frac{\sin x}{\cos x} = \frac{1 - \sin x}{\cos x} \]

### Step 2: Identify the limit form
Now, we need to evaluate the limit:
\[ \lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{\cos x} \]
As \(x\) approaches \(\frac{\pi}{2}\), \(\sin x\) approaches \(1\) and \(\cos x\) approaches \(0\). This gives us the indeterminate form \(\frac{0}{0}\).

### Step 3: Apply L'Hôpital's Rule
Since we have the indeterminate form \(\frac{0}{0}\), we can apply L'Hôpital's Rule, which states that:
\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]
if the limit on the right exists. Here, \(f(x) = 1 - \sin x\) and \(g(x) = \cos x\).

### Step 4: Differentiate the numerator and denominator
Now we differentiate \(f(x)\) and \(g(x)\):
- The derivative of \(f(x) = 1 - \sin x\) is \(f'(x) = -\cos x\).
- The derivative of \(g(x) = \cos x\) is \(g'(x) = -\sin x\).

### Step 5: Rewrite the limit using derivatives
Now we can rewrite the limit:
\[ \lim_{x \to \frac{\pi}{2}} \frac{1 - \sin x}{\cos x} = \lim_{x \to \frac{\pi}{2}} \frac{-\cos x}{-\sin x} = \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{\sin x} \]

### Step 6: Evaluate the new limit
Now we evaluate:
\[ \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{\sin x} \]
As \(x\) approaches \(\frac{\pi}{2}\), \(\cos x\) approaches \(0\) and \(\sin x\) approaches \(1\). Thus, we have:
\[ \lim_{x \to \frac{\pi}{2}} \frac{\cos x}{\sin x} = \frac{0}{1} = 0 \]

### Conclusion
Therefore, we conclude that:
\[ \lim_{x \to \frac{\pi}{2}} (\sec x - \tan x) = 0 \]