Show that `Lim_(x to pi/2)` tan x does not exist .

Answer

To show that \(\lim_{x \to \frac{\pi}{2}} \tan x\) does not exist, we can follow these steps:

### Step 1: Understand the definition of the tangent function
The tangent function is defined as:
\[ \tan x = \frac{\sin x}{\cos x} \]

### Step 2: Substitute the limit into the tangent function
We want to find:
\[ \lim_{x \to \frac{\pi}{2}} \tan x = \lim_{x \to \frac{\pi}{2}} \frac{\sin x}{\cos x} \]

### Step 3: Evaluate the sine and cosine at \(\frac{\pi}{2}\)
As \(x\) approaches \(\frac{\pi}{2}\):
- \(\sin\left(\frac{\pi}{2}\right) = 1\)
- \(\cos\left(\frac{\pi}{2}\right) = 0\)

### Step 4: Analyze the limit
Substituting these values into the limit gives:
\[ \lim_{x \to \frac{\pi}{2}} \frac{\sin x}{\cos x} = \frac{1}{0} \]
Since division by zero is undefined, we need to analyze the behavior of \(\tan x\) as \(x\) approaches \(\frac{\pi}{2}\) from both sides.

### Step 5: Check the left-hand limit
As \(x\) approaches \(\frac{\pi}{2}\) from the left (\(x \to \frac{\pi}{2}^-\)):
- \(\cos x\) approaches \(0\) and is positive, hence \(\tan x\) approaches \(+\infty\).

### Step 6: Check the right-hand limit
As \(x\) approaches \(\frac{\pi}{2}\) from the right (\(x \to \frac{\pi}{2}^+\)):
- \(\cos x\) approaches \(0\) and is negative, hence \(\tan x\) approaches \(-\infty\).

### Step 7: Conclusion
Since the left-hand limit approaches \(+\infty\) and the right-hand limit approaches \(-\infty\), we conclude that:
\[ \lim_{x \to \frac{\pi}{2}} \tan x \text{ does not exist.} \]