Evaluating the Limit of $ \frac{x^{3} - 1}{x - 1} $ as $ x $ approaches $ 1 $

The question asks us to find the value of the limit:

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

If we try to substitute $ x = 1 $ directly into the expression, we get:

\[ \frac{1^{3} - 1}{1 - 1} = \frac{1 - 1}{1 - 1} = \frac{0}{0} \]

This is an indeterminate form, which means we need to use other methods to evaluate the limit. Here are two common methods:

Method 1: Using Factorization

The numerator, $ x^3 - 1 $, is a difference of cubes. The formula for the difference of cubes is $ a^3 - b^3 = (a - b)(a^2 + ab + b^2) $. In our case, $ a = x $ and $ b = 1 $.

So, we can factor the numerator as:

\[ x^3 - 1 = (x - 1)(x^2 + x \cdot 1 + 1^2) = (x - 1)(x^2 + x + 1) \]

Now, substitute this factorization back into the limit expression:

\[ \lim_{x \to 1} \frac{(x - 1)(x^2 + x + 1)}{x - 1} \]

Since $ x $ approaches $ 1 $, $ x $ is not exactly equal to $ 1 $, so $ x - 1 \neq 0 $. We can cancel the $ (x - 1) $ term from the numerator and the denominator:

\[ \lim_{x \to 1} (x^2 + x + 1) \]

Now, we can substitute $ x = 1 $ into the simplified expression:

\[ 1^2 + 1 + 1 = 1 + 1 + 1 = 3 \]

Method 2: Using L'Hôpital's Rule

Since we obtained the indeterminate form $ \frac{0}{0} $ upon direct substitution, we can apply L'Hôpital's Rule. This rule states that if $ \lim_{x \to c} \frac{f(x)}{g(x)} $ results in an indeterminate form ($ \frac{0}{0} $ or $ \frac{\infty}{\infty} $), then the limit is equal to $ \lim_{x \to c} \frac{f'(x)}{g'(x)} $, provided the latter limit exists.

Let $ f(x) = x^3 - 1 $ and $ g(x) = x - 1 $.

Find the derivatives:

• The derivative of the numerator is $ f'(x) = \frac{d}{dx}(x^3 - 1) = 3x^2 $.
• The derivative of the denominator is $ g'(x) = \frac{d}{dx}(x - 1) = 1 $.

Now, apply L'Hôpital's Rule:

\[ \lim_{x \to 1} \frac{f'(x)}{g'(x)} = \lim_{x \to 1} \frac{3x^2}{1} \]

Substitute $ x = 1 $ into the new expression:

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

Conclusion

Both the factorization method and L'Hôpital's Rule show that the limit of the given function as $ x $ approaches $ 1 $ is $ 3 $. Therefore, the value of the limit is $ 3 $.