If `f(x)=(2)/(x-3),g(x)=(x-3)/(x+4)," and "h(x)=-(2(2x+1))/(x^(2)+x-12)," then "lim_(xto3) [f(x)+g(x)+h(x)]" is "`

Options

A. 1
B. `oo`
C. `sqrt(2)`
D. none of these

Explanation

To solve the limit problem step by step, we will evaluate the limit as \( x \) approaches 3 for the expression \( f(x) + g(x) + h(x) \).

### Step 1: Define the functions
We have the following functions:
- \( f(x) = \frac{2}{x - 3} \)
- \( g(x) = \frac{x - 3}{x + 4} \)
- \( h(x) = -\frac{2(2x + 1)}{x^2 + x - 12} \)

### Step 2: Substitute the functions into the limit
We need to find:
\[ \lim_{x \to 3} \left( f(x) + g(x) + h(x) \right) \]
This becomes:
\[ \lim_{x \to 3} \left( \frac{2}{x - 3} + \frac{x - 3}{x + 4} - \frac{2(2x + 1)}{x^2 + x - 12} \right) \]

### Step 3: Factor the denominator of \( h(x) \)
The denominator of \( h(x) \) can be factored:
\[ x^2 + x - 12 = (x - 3)(x + 4) \]
Thus, we can rewrite \( h(x) \):
\[ h(x) = -\frac{2(2x + 1)}{(x - 3)(x + 4)} \]

### Step 4: Combine the functions over a common denominator
The common denominator for \( f(x) \), \( g(x) \), and \( h(x) \) is \( (x - 3)(x + 4) \). We rewrite the limit:
\[ \lim_{x \to 3} \left( \frac{2(x + 4)}{(x - 3)(x + 4)} + \frac{(x - 3)(x - 3)}{(x - 3)(x + 4)} - \frac{2(2x + 1)}{(x - 3)(x + 4)} \right) \]

### Step 5: Combine the numerators
Combining the numerators gives:
\[ \lim_{x \to 3} \frac{2(x + 4) + (x - 3)^2 - 2(2x + 1)}{(x - 3)(x + 4)} \]

### Step 6: Expand the numerator
Expanding the numerator:
- \( 2(x + 4) = 2x + 8 \)
- \( (x - 3)^2 = x^2 - 6x + 9 \)
- \( -2(2x + 1) = -4x - 2 \)

Combining these:
\[ 2x + 8 + x^2 - 6x + 9 - 4x - 2 = x^2 - 8x + 15 \]

### Step 7: Factor the numerator
Now we have:
\[ \lim_{x \to 3} \frac{x^2 - 8x + 15}{(x - 3)(x + 4)} \]
Factoring \( x^2 - 8x + 15 \):
\[ x^2 - 8x + 15 = (x - 3)(x - 5) \]

### Step 8: Simplify the expression
Thus, we can simplify:
\[ \lim_{x \to 3} \frac{(x - 3)(x - 5)}{(x - 3)(x + 4)} \]
Cancelling \( (x - 3) \):
\[ \lim_{x \to 3} \frac{x - 5}{x + 4} \]

### Step 9: Evaluate the limit
Now we can substitute \( x = 3 \):
\[ \frac{3 - 5}{3 + 4} = \frac{-2}{7} \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 3} \left( f(x) + g(x) + h(x) \right) = -\frac{2}{7} \]