Find the value of `a ,b` if `f(x) = {((a e^(1//|x+2|) - 1)/(2-e^(1//|x+2|)),; -3 lt x lt -2), (b,; x = -2), (sin((x^4-16)/(x^5+32)),; -2 lt x lt 0):}` is continuous at `x = -2.`

Options

A. ` a=sin "" (2)/(5)`
B. `b=-sin ""(2)/(5)`
C. ` a=- sin "" (1)/(5)`
D. `b=sin "" (1)/(5)`

Answer

To find the values of \( a \) and \( b \) such that the function

\[ f(x) = \begin{cases} \frac{a e^{\frac{1}{|x+2|}} - 1}{2 - e^{\frac{1}{|x+2|}} & \text{for } -3 < x < -2 \\ b & \text{for } x = -2 \\ \sin\left(\frac{x^4 - 16}{x^5 + 32}\right) & \text{for } -2 < x < 0 \end{cases} \]

is continuous at \( x = -2 \), we need to ensure that the left-hand limit (LHL) and right-hand limit (RHL) at \( x = -2 \) equal \( f(-2) \).

### Step 1: Calculate the Left-Hand Limit (LHL) as \( x \to -2^- \)

The LHL is given by:

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

As \( x \to -2^- \), \( |x+2| = -(x+2) \) and approaches \( 0^+ \). Thus, we can substitute \( h = -2 - x \) which implies \( h \to 0^+ \) as \( x \to -2^- \). Therefore:

\[ \lim_{h \to 0^+} \frac{a e^{\frac{1}{h}} - 1}{2 - e^{\frac{1}{h}}} \]

### Step 2: Simplifying the Limit

As \( h \to 0^+ \), \( e^{\frac{1}{h}} \to \infty \). Thus, we can rewrite the limit:

\[ \lim_{h \to 0^+} \frac{a e^{\frac{1}{h}} - 1}{2 - e^{\frac{1}{h}}} = \lim_{h \to 0^+} \frac{a e^{\frac{1}{h}}}{-e^{\frac{1}{h}}} = \lim_{h \to 0^+} -a \]

This gives us:

\[ \text{LHL} = -a \]

### Step 3: Calculate the Right-Hand Limit (RHL) as \( x \to -2^+ \)

The RHL is given by:

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

### Step 4: Simplifying the RHL

We can factor the numerator and denominator:

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

Thus, we have:

\[ \lim_{x \to -2^+} \frac{(x+2)(x^2 - 4)(x^2 + 4)}{(x+2)(x^4 - 2x^3 + 4x^2 - 8x + 16)} \]

Cancelling \( (x + 2) \):

\[ \lim_{x \to -2^+} \frac{(x^2 - 4)(x^2 + 4)}{x^4 - 2x^3 + 4x^2 - 8x + 16} \]

Evaluating this limit gives:

\[ \frac{(-2^2 - 4)(-2^2 + 4)}{-2^4 + 2(-2)^3 + 4(-2)^2 - 8(-2) + 16} = \frac{(0)(0)}{0} = \text{Indeterminate} \]

Using L'Hôpital's Rule or further simplification, we find:

\[ \text{RHL} = -\frac{8}{-16} = \frac{1}{2} \]

### Step 5: Setting the Limits Equal

For continuity at \( x = -2 \):

\[ -a = \frac{1}{2} \quad \text{and} \quad b = \frac{1}{2} \]

### Step 6: Solving for \( a \) and \( b \)

From \( -a = \frac{1}{2} \):

\[ a = -\frac{1}{2} \]

And from \( b = \frac{1}{2} \):

\[ b = \frac{1}{2} \]

### Final Values

Thus, the values of \( a \) and \( b \) are:

\[ \boxed{a = -\frac{1}{2}, b = \frac{1}{2}} \]