Evaluate : `lim_(xrarr0)((1-x)^(n)-1)/(x)`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{(1 - x)^n - 1}{x}, \]

we can follow these steps:

### Step 1: Substitute \( x = 0 \)

First, we substitute \( x = 0 \) into the expression:

\[ \frac{(1 - 0)^n - 1}{0} = \frac{1 - 1}{0} = \frac{0}{0}. \]

Since we get an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule.

**Hint:** If you encounter \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) when evaluating a limit, consider using L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule

According to L'Hôpital's Rule, we differentiate the numerator and the denominator separately:

\[ \lim_{x \to 0} \frac{(1 - x)^n - 1}{x} = \lim_{x \to 0} \frac{\frac{d}{dx}((1 - x)^n - 1)}{\frac{d}{dx}(x)}. \]

### Step 3: Differentiate the Numerator and Denominator

Now we differentiate the numerator:

1. The derivative of \((1 - x)^n\) is \(n(1 - x)^{n-1} \cdot (-1) = -n(1 - x)^{n-1}\).
2. The derivative of \(-1\) is \(0\).
3. The derivative of \(x\) is \(1\).

Thus, we have:

\[ \lim_{x \to 0} \frac{-n(1 - x)^{n-1}}{1}. \]

### Step 4: Substitute \( x = 0 \) Again

Now, we substitute \( x = 0 \) into the differentiated limit:

\[ -n(1 - 0)^{n-1} = -n(1)^{n-1} = -n. \]

### Conclusion

Therefore, the limit evaluates to:

\[ \lim_{x \to 0} \frac{(1 - x)^n - 1}{x} = -n. \]

**Final Answer:** \(-n\)

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