Value of the `underset(x to -2)lim (tan pi x)/(x+2)+underset(x to oo)lim (1+1/x^(2))^(x)` is

Options

A. `pi+1`
B. `pi-1`
C. `pi//2+1`
D. `pi//3-1`

Answer

To solve the given problem, we need to evaluate two limits and then add their results together.

### Step 1: Evaluate the first limit
We need to find:
\[ \lim_{x \to -2} \frac{\tan(\pi x)}{x + 2} \]
Substituting \(x = -2\):
\[ \tan(\pi \cdot -2) = \tan(-2\pi) = 0 \]
And:
\[ x + 2 = -2 + 2 = 0 \]
This gives us the indeterminate form \(\frac{0}{0}\). Therefore, we apply L'Hôpital's Rule, which states that if we have an indeterminate form \(\frac{0}{0}\), we can take the derivative of the numerator and the derivative of the denominator.

### Step 2: Differentiate the numerator and denominator
Differentiating the numerator:
\[ \frac{d}{dx}[\tan(\pi x)] = \pi \sec^2(\pi x) \]
Differentiating the denominator:
\[ \frac{d}{dx}[x + 2] = 1 \]

### Step 3: Apply L'Hôpital's Rule
Now we can apply L'Hôpital's Rule:
\[ \lim_{x \to -2} \frac{\tan(\pi x)}{x + 2} = \lim_{x \to -2} \frac{\pi \sec^2(\pi x)}{1} \]
Substituting \(x = -2\):
\[ \sec^2(-2\pi) = \sec^2(2\pi) = 1 \]
Thus:
\[ \lim_{x \to -2} \frac{\tan(\pi x)}{x + 2} = \pi \cdot 1 = \pi \]

### Step 4: Evaluate the second limit
Now we need to find:
\[ \lim_{x \to \infty} \left(1 + \frac{1}{x^2}\right)^x \]
As \(x\) approaches infinity, \(\frac{1}{x^2}\) approaches 0, so we have:
\[ \lim_{x \to \infty} \left(1 + 0\right)^x = 1^\infty \]
This is another indeterminate form. We can rewrite this limit 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:
\[ x \ln\left(1 + \frac{1}{x^2}\right) \approx x \cdot \frac{1}{x^2} = \frac{1}{x} \]
As \(x\) approaches infinity, \(\frac{1}{x}\) approaches 0. Therefore:
\[ \lim_{x \to \infty} x \ln\left(1 + \frac{1}{x^2}\right) = 0 \]
This gives us:
\[ \lim_{x \to \infty} \left(1 + \frac{1}{x^2}\right)^x = e^0 = 1 \]

### Step 6: Combine the results
Now we can add the results of the two limits:
\[ \lim_{x \to -2} \frac{\tan(\pi x)}{x + 2} + \lim_{x \to \infty} \left(1 + \frac{1}{x^2}\right)^x = \pi + 1 \]

### Final Answer
Thus, the final answer is:
\[ \pi + 1 \]