Evaluate the following limits: `lim_(xrarr3)((x^(3)-27)/(x-3))`

Answer

To evaluate the limit

\[ \lim_{x \to 3} \frac{x^3 - 27}{x - 3}, \]

we will follow these steps:

### Step 1: Direct Substitution
First, we try to substitute \( x = 3 \) directly into the expression:

\[ \frac{3^3 - 27}{3 - 3} = \frac{27 - 27}{0} = \frac{0}{0}. \]

This results in an indeterminate form \( \frac{0}{0} \).

**Hint:** If direct substitution leads to an indeterminate form, consider using L'Hôpital's Rule or algebraic manipulation.

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if we have \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can take the derivative of the numerator and the derivative of the denominator.

Calculating the derivatives:

- The derivative of the numerator \( x^3 - 27 \) is \( 3x^2 \).
- The derivative of the denominator \( x - 3 \) is \( 1 \).

Now we rewrite the limit:

\[ \lim_{x \to 3} \frac{3x^2}{1}. \]

**Hint:** Remember that L'Hôpital's Rule can be applied multiple times if you still encounter an indeterminate form.

### Step 3: Substitute Again
Now we substitute \( x = 3 \) into the new limit:

\[ \frac{3(3^2)}{1} = \frac{3 \cdot 9}{1} = 27. \]

**Hint:** After applying L'Hôpital's Rule, always check if substituting the limit gives a determinate value.

### Final Answer
Thus, the limit is

\[ \lim_{x \to 3} \frac{x^3 - 27}{x - 3} = 27. \]