Evaluate the following limits :
`lim_(n to 2)(n^(3)-8)/(n^(2)-4)`

Answer

To evaluate the limit \( \lim_{n \to 2} \frac{n^3 - 8}{n^2 - 4} \), we can follow these steps:

### Step 1: Identify the form of the limit
First, substitute \( n = 2 \) into the expression:
\[ \frac{2^3 - 8}{2^2 - 4} = \frac{8 - 8}{4 - 4} = \frac{0}{0} \]
This indicates that we have an indeterminate form \( \frac{0}{0} \).

**Hint:** When you encounter \( \frac{0}{0} \), consider factoring the numerator and denominator.

### Step 2: Factor the numerator and denominator
The numerator \( n^3 - 8 \) can be factored as a difference of cubes:
\[ n^3 - 8 = n^3 - 2^3 = (n - 2)(n^2 + 2n + 4) \]
The denominator \( n^2 - 4 \) can be factored as a difference of squares:
\[ n^2 - 4 = n^2 - 2^2 = (n - 2)(n + 2) \]

### Step 3: Rewrite the limit with factored forms
Now, substitute the factored forms back into the limit:
\[ \lim_{n \to 2} \frac{(n - 2)(n^2 + 2n + 4)}{(n - 2)(n + 2)} \]

**Hint:** Look for common factors in the numerator and denominator that can be canceled.

### Step 4: Cancel the common factors
We can cancel the \( (n - 2) \) terms from the numerator and denominator:
\[ \lim_{n \to 2} \frac{n^2 + 2n + 4}{n + 2} \]

### Step 5: Substitute \( n = 2 \) again
Now substitute \( n = 2 \) into the simplified expression:
\[ \frac{2^2 + 2 \cdot 2 + 4}{2 + 2} = \frac{4 + 4 + 4}{4} = \frac{12}{4} = 3 \]

### Final Answer
Thus, the limit is:
\[ \lim_{n \to 2} \frac{n^3 - 8}{n^2 - 4} = 3 \]

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