If `A=lim_(xrarr-2)(tanpix)/(x+2)+lim_(xrarroo)(1+(1)/(x^(2)))^(x)`, then

Options

A. `A gt 3`
B. `A gt 4`
C. `A lt 4`
D. A is a transcedental number

Answer

To solve the problem, we need to evaluate the expression:

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

### Step 1: Evaluate the first limit

We start with the limit:

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

Substituting \( x = -2 \) directly gives us:

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

This results in the indeterminate form \( \frac{0}{0} \). Therefore, we can apply L'Hôpital's Rule.

### Step 2: 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:

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

### Step 3: Substitute \( x = -2 \)

Now we substitute \( x = -2 \):

\[ \sec^2(-2\pi) = \sec^2(2\pi) = \frac{1}{\cos^2(2\pi)} = \frac{1}{1^2} = 1 \]

Thus, we get:

\[ \lim_{x \to -2} \frac{\tan(\pi x)}{x + 2} = \pi \cdot 1 = \pi \]

### Step 4: Evaluate the second limit

Next, we evaluate the second limit:

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

This limit is of the form \( 1^\infty \). We can rewrite it using the exponential function:

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

### Step 5: Simplify the logarithm

Using the Taylor expansion \( \ln(1 + u) \approx u \) for small \( u \):

\[ \ln\left(1 + \frac{1}{x^2}\right) \approx \frac{1}{x^2} \]

Thus,

\[ x \ln\left(1 + \frac{1}{x^2}\right) \approx x \cdot \frac{1}{x^2} = \frac{1}{x} \]

### Step 6: Evaluate the limit

Now we evaluate:

\[ \lim_{x \to \infty} \frac{1}{x} = 0 \]

So we have:

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

### Step 7: Combine the results

Now we can combine both limits:

\[ A = \pi + 1 \]

### Final Result

Thus, the final value of \( A \) is:

\[ A = \pi + 1 \]