If `l=lim_(xto-2) (tanpix)/(x+2)+lim_(xtooo) ( (1+1)/(x^2)^2)`, then which one of the following is not correct?

Options

A. `lgt 3`
B. `lgt4`
C. `l lt 4`
D. l is a transcendental number

Answer

To solve the problem, we need to evaluate the expression for \( L \):

\[ L = \lim_{x \to -2} \frac{\tan(\pi x)}{x + 2} + \lim_{x \to \infty} \left( \frac{1 + 1}{(x^2)^2} \right) \]

### Step 1: Evaluate the second limit
First, we evaluate the limit as \( x \) approaches infinity:

\[ \lim_{x \to \infty} \left( \frac{1 + 1}{(x^2)^2} \right) = \lim_{x \to \infty} \left( \frac{2}{x^4} \right) \]

As \( x \) approaches infinity, \( x^4 \) approaches infinity, thus:

\[ \frac{2}{x^4} \to 0 \]

So,

\[ \lim_{x \to \infty} \left( \frac{1 + 1}{(x^2)^2} \right) = 0 \]

### Step 2: Evaluate the first limit
Next, we evaluate the limit as \( x \) approaches -2:

\[ \lim_{x \to -2} \frac{\tan(\pi x)}{x + 2} \]

Substituting \( x = -2 \):

\[ \tan(-2\pi) = 0 \quad \text{and} \quad x + 2 = 0 \]

This gives us a \( \frac{0}{0} \) indeterminate form, so we apply L'Hôpital's Rule:

\[ L = \lim_{x \to -2} \frac{\tan(\pi x)}{x + 2} \]

### Step 3: Apply L'Hôpital's Rule
Using L'Hôpital's Rule, we differentiate the numerator and denominator:

1. The derivative of the numerator \( \tan(\pi x) \) is \( \pi \sec^2(\pi x) \).
2. The derivative of the denominator \( x + 2 \) is \( 1 \).

Thus, we have:

\[ L = \lim_{x \to -2} \frac{\pi \sec^2(\pi x)}{1} \]

Substituting \( x = -2 \):

\[ \sec^2(-2\pi) = \sec^2(0) = 1 \]

So,

\[ L = \pi \cdot 1 = \pi \]

### Step 4: Combine the limits
Now we combine both limits:

\[ L = \pi + 0 = \pi \]

### Step 5: Analyze the options
Now we need to check which of the following statements is not correct:

1. \( L > 3 \) (True, since \( \pi \approx 3.14 \))
2. \( L > 4 \) (False, since \( \pi < 4 \))
3. \( L < 4 \) (True, since \( \pi < 4 \))
4. \( L \) is a transcendental number (True, since \( \pi \) is transcendental)

Thus, the statement that is not correct is:

**\( L > 4 \)**.