If `lim_(xto-2^(-)) (ae^(1//|x+2|)-1)/(2-e^(1//|x+2|))=lim_(xto-2^(+)) sin((x^(4)-16)/(x^(5)+32)),` then a is

Options

A. 1
B. -1
C. 0
D. none of these

Explanation

To solve the given limit problem, we need to find the value of \( a \) such that:

\[ \lim_{x \to -2^-} \frac{a e^{\frac{1}{|x+2|}} - 1}{2 - e^{\frac{1}{|x+2|}}} = \lim_{x \to -2^+} \sin\left(\frac{x^4 - 16}{x^5 + 32}\right) \]

### Step 1: Evaluate the left-hand limit

As \( x \to -2^- \), \( |x+2| = -(x+2) \) and approaches \( 0 \) from the negative side. Thus, we can rewrite the limit:

\[ \lim_{x \to -2^-} \frac{a e^{\frac{1}{|x+2|}} - 1}{2 - e^{\frac{1}{|x+2|}}} \]

As \( x \to -2^- \), \( e^{\frac{1}{|x+2|}} \to e^{+\infty} \) (since \( \frac{1}{|x+2|} \to +\infty \)). Therefore, we have:

\[ \lim_{x \to -2^-} \frac{a \cdot \infty - 1}{2 - \infty} = \frac{\infty}{-\infty} \]

This is an indeterminate form, so we can apply L'Hôpital's rule.

### Step 2: Apply L'Hôpital's Rule

Differentiating the numerator and denominator with respect to \( x \):

- The derivative of the numerator \( a e^{\frac{1}{|x+2|}} \) is \( a e^{\frac{1}{|x+2|}} \cdot \left(-\frac{1}{(x+2)^2}\right) \) (using the chain rule).
- The derivative of the denominator \( 2 - e^{\frac{1}{|x+2|}} \) is \( -e^{\frac{1}{|x+2|}} \cdot \left(-\frac{1}{(x+2)^2}\right) \).

Thus, we have:

\[ \lim_{x \to -2^-} \frac{a e^{\frac{1}{|x+2|}} \cdot \left(-\frac{1}{(x+2)^2}\right)}{-e^{\frac{1}{|x+2|}} \cdot \left(-\frac{1}{(x+2)^2}\right)} = \lim_{x \to -2^-} \frac{a}{1} = a \]

### Step 3: Evaluate the right-hand limit

Now, we evaluate the right-hand limit:

\[ \lim_{x \to -2^+} \sin\left(\frac{x^4 - 16}{x^5 + 32}\right) \]

As \( x \to -2^+ \):

- \( x^4 - 16 = (x^2 - 4)(x^2 + 4) = (x-2)(x+2)(x^2 + 4) \)
- \( x^5 + 32 = (x + 2)(x^4 - 2x^3 + 4x^2 - 8x + 16) \)

Both the numerator and denominator approach \( 0 \), giving us the form \( \frac{0}{0} \).

### Step 4: Simplify the right-hand limit

Using L'Hôpital's Rule again, we differentiate the numerator and denominator:

1. Differentiate the numerator \( x^4 - 16 \) to get \( 4x^3 \).
2. Differentiate the denominator \( x^5 + 32 \) to get \( 5x^4 \).

Thus, we have:

\[ \lim_{x \to -2^+} \frac{4x^3}{5x^4} \]

Substituting \( x = -2 \):

\[ = \frac{4(-2)^3}{5(-2)^4} = \frac{4(-8)}{5(16)} = \frac{-32}{80} = -\frac{2}{5} \]

### Step 5: Set the limits equal

Now we set the two limits equal to each other:

\[ a = -\frac{2}{5} \]

### Final Answer

Thus, the value of \( a \) is:

\[ \boxed{-\frac{2}{5}} \]