Match the following :
`{:("Column-I" , " Column - II"), ("(A)" lim_(x to pi/4)(sin2x)^(tan^(2)2x), "(p)" 1/2),("(B)" lim_(x to oo) ((2x-1)/(2x+1))^(x) , "(q)"e^(-1/2)),("(C)" lim_(x to pi/2)(tanx)^(tan 2x) , "(r)" e^(-1)),("(D)" lim_(x to pi/4) tan2x tan(pi/4-x), "(s)" 1):}`

Answer

To solve the problem of matching the limits from Column-I with their corresponding values in Column-II, we will analyze each limit step by step.

### Step 1: Evaluate Limit A
**Limit A:**
\[ \lim_{x \to \frac{\pi}{4}} (\sin 2x)^{\tan^2 2x} \]

1. Substitute \(x = \frac{\pi}{4}\):
\[ \sin 2\left(\frac{\pi}{4}\right) = \sin \frac{\pi}{2} = 1 \]
\[ \tan 2\left(\frac{\pi}{4}\right) = \tan \frac{\pi}{2} \text{ (undefined, but we analyze the limit)} \]

2. This gives us the form \(1^{\infty}\). We can use the logarithmic limit:
\[ y = (\sin 2x)^{\tan^2 2x} \implies \ln y = \tan^2 2x \cdot \ln(\sin 2x) \]

3. As \(x \to \frac{\pi}{4}\), \(\tan^2 2x \to \infty\) and \(\ln(\sin 2x) \to 0\). We rewrite:
\[ \lim_{x \to \frac{\pi}{4}} \tan^2 2x \cdot \ln(\sin 2x) \]

4. Using L'Hôpital's Rule or Taylor expansion, we find:
\[ \lim_{x \to \frac{\pi}{4}} \tan^2 2x \cdot \ln(\sin 2x) = -\frac{1}{2} \]

5. Therefore,
\[ \lim_{x \to \frac{\pi}{4}} (\sin 2x)^{\tan^2 2x} = e^{-\frac{1}{2}}. \]
This corresponds to option **(q)**.

### Step 2: Evaluate Limit B
**Limit B:**
\[ \lim_{x \to \infty} \left(\frac{2x-1}{2x+1}\right)^{x} \]

1. Rewrite the expression:
\[ \frac{2x-1}{2x+1} = 1 - \frac{2}{2x+1} \]

2. As \(x \to \infty\), this approaches \(1\), leading to the form \(1^{\infty}\).

3. Using the logarithmic limit:
\[ \ln y = x \ln\left(1 - \frac{2}{2x+1}\right) \approx -\frac{2x}{2x+1} \text{ (using Taylor expansion)} \]

4. This simplifies to:
\[ \lim_{x \to \infty} -\frac{2x}{2x+1} = -1 \]

5. Therefore,
\[ \lim_{x \to \infty} \left(\frac{2x-1}{2x+1}\right)^{x} = e^{-1}. \]
This corresponds to option **(r)**.

### Step 3: Evaluate Limit C
**Limit C:**
\[ \lim_{x \to \frac{\pi}{2}} (\tan x)^{\tan 2x} \]

1. Substitute \(x = \frac{\pi}{2}\):
\[ \tan\left(\frac{\pi}{2}\right) \to \infty \quad \text{and} \quad \tan(2x) = \tan(\pi) = 0 \]

2. This gives us the form \(\infty^0\). We rewrite:
\[ y = (\tan x)^{\tan 2x} \implies \ln y = \tan 2x \cdot \ln(\tan x) \]

3. As \(x \to \frac{\pi}{2}\), \(\tan 2x \to 0\) and \(\ln(\tan x) \to \infty\). We need to analyze:
\[ \lim_{x \to \frac{\pi}{2}} \tan 2x \cdot \ln(\tan x) \]

4. Using L'Hôpital's Rule, we find:
\[ \lim_{x \to \frac{\pi}{2}} \tan 2x \cdot \ln(\tan x) = 1 \]

5. Therefore,
\[ \lim_{x \to \frac{\pi}{2}} (\tan x)^{\tan 2x} = e^{0} = 1. \]
This corresponds to option **(s)**.

### Step 4: Evaluate Limit D
**Limit D:**
\[ \lim_{x \to \frac{\pi}{4}} \tan 2x \tan\left(\frac{\pi}{4} - x\right) \]

1. Substitute \(x = \frac{\pi}{4}\):
\[ \tan 2\left(\frac{\pi}{4}\right) = \tan \frac{\pi}{2} \text{ (undefined)} \]

2. Rewrite:
\[ \tan 2x = \frac{2\tan x}{1 - \tan^2 x} \quad \text{and} \quad \tan\left(\frac{\pi}{4} - x\right) = \frac{1 - \tan x}{1 + \tan x} \]

3. As \(x \to \frac{\pi}{4}\), both terms approach \(1\).

4. Therefore,
\[ \lim_{x \to \frac{\pi}{4}} \tan 2x \tan\left(\frac{\pi}{4} - x\right) = 1 \cdot 1 = 1. \]
This corresponds to option **(p)**.

### Final Matching
- (A) → (q)
- (B) → (r)
- (C) → (s)
- (D) → (p)