If `lim_(xto3)(x^(3)+cx^(2)+5x+12)/(x^(2)-7x+12)=l` (finite real number), ten

Options

A. `l=4`
B. `c=-6`
C. `c=4`
D. `x epsilonR`

Answer

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

\[ \lim_{x \to 3} \frac{x^3 + cx^2 + 5x + 12}{x^2 - 7x + 12} = l \]

is a finite real number.

### Step 1: Evaluate the limit at \( x = 3 \)

First, we substitute \( x = 3 \) into the denominator:

\[ x^2 - 7x + 12 = 3^2 - 7 \cdot 3 + 12 = 9 - 21 + 12 = 0 \]

Since the denominator is zero, we need to check the numerator:

\[ x^3 + cx^2 + 5x + 12 = 3^3 + c \cdot 3^2 + 5 \cdot 3 + 12 = 27 + 9c + 15 + 12 = 54 + 9c \]

For the limit to be finite, the numerator must also equal zero when \( x = 3 \):

\[ 54 + 9c = 0 \]

### Step 2: Solve for \( c \)

Now, we solve for \( c \):

\[ 9c = -54 \implies c = -6 \]

### Step 3: Substitute \( c \) back into the limit

Now that we have \( c = -6 \), we substitute this value back into the original limit:

\[ \lim_{x \to 3} \frac{x^3 - 6x^2 + 5x + 12}{x^2 - 7x + 12} \]

### Step 4: Differentiate using L'Hôpital's Rule

Since we still have a \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule:

Differentiate the numerator and denominator:

- Numerator:
\[ \frac{d}{dx}(x^3 - 6x^2 + 5x + 12) = 3x^2 - 12x + 5 \]

- Denominator:
\[ \frac{d}{dx}(x^2 - 7x + 12) = 2x - 7 \]

### Step 5: Evaluate the limit again

Now we evaluate the limit:

\[ \lim_{x \to 3} \frac{3x^2 - 12x + 5}{2x - 7} \]

Substituting \( x = 3 \):

Numerator:
\[ 3(3^2) - 12(3) + 5 = 27 - 36 + 5 = -4 \]

Denominator:
\[ 2(3) - 7 = 6 - 7 = -1 \]

Thus, the limit becomes:

\[ \frac{-4}{-1} = 4 \]

### Final Result

The value of \( c \) is \( -6 \) and the limit \( l \) is \( 4 \).

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