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

Options

A. `-2`
B. `-1`
C. `-2/7`
D. `0`

Answer

To solve the limit \( \lim_{x \to 3} [f(x) + g(x) + h(x)] \), where \( f(x) = \frac{2}{x-3} \), \( g(x) = \frac{x-3}{x+4} \), and \( h(x) = -\frac{2(2x+1)}{x^2+x-12} \), we will follow these steps:

### Step 1: Analyze the functions at \( x = 3 \)
First, we substitute \( x = 3 \) into each function to see if we get any indeterminate forms.

- For \( f(3) = \frac{2}{3-3} \) → undefined (infinity)
- For \( g(3) = \frac{3-3}{3+4} = \frac{0}{7} = 0 \)
- For \( h(3) \): We need to factor the denominator.

### Step 2: Factor the denominator of \( h(x) \)
The denominator of \( h(x) \) is \( x^2 + x - 12 \). We can factor this:

\[ x^2 + x - 12 = (x - 3)(x + 4) \]

Thus, we can rewrite \( h(x) \):

\[ h(x) = -\frac{2(2x + 1)}{(x - 3)(x + 4)} \]

### Step 3: Rewrite the limit expression
Now we can express the limit as:

\[ \lim_{x \to 3} \left[ \frac{2}{x-3} + \frac{x-3}{x+4} - \frac{2(2x+1)}{(x-3)(x+4)} \right] \]

### Step 4: Combine the terms
To combine these fractions, we will find a common denominator, which is \( (x-3)(x+4) \):

\[ \frac{2(x+4)}{(x-3)(x+4)} + \frac{(x-3)(x-3)}{(x-3)(x+4)} - \frac{2(2x+1)}{(x-3)(x+4)} \]

This simplifies to:

\[ \frac{2(x+4) + (x-3)^2 - 2(2x+1)}{(x-3)(x+4)} \]

### Step 5: Simplify the numerator
Now we simplify the numerator:

1. Expand \( (x-3)^2 = x^2 - 6x + 9 \)
2. Combine all terms:

\[ 2(x + 4) = 2x + 8 \]
\[ -2(2x + 1) = -4x - 2 \]

Putting it all together:

\[ 2x + 8 + (x^2 - 6x + 9) - (4x + 2) = x^2 + (2x - 6x - 4x) + (8 + 9 - 2) = x^2 - 8x + 15 \]

### Step 6: Factor the numerator
Now we factor \( x^2 - 8x + 15 \):

\[ x^2 - 8x + 15 = (x - 3)(x - 5) \]

### Step 7: Cancel common factors
Now we can cancel \( (x - 3) \):

\[ \lim_{x \to 3} \frac{(x - 3)(x - 5)}{(x - 3)(x + 4)} = \lim_{x \to 3} \frac{x - 5}{x + 4} \]

### Step 8: Substitute \( x = 3 \)
Now we substitute \( x = 3 \):

\[ \frac{3 - 5}{3 + 4} = \frac{-2}{7} \]

### Final Answer
Thus, the limit is:

\[ \lim_{x \to 3} [f(x) + g(x) + h(x)] = -\frac{2}{7} \]