Question

The value of \( \displaystyle \lim_{x \to 0} \frac{(1-x)^n - 1}{x} \) is

• A. \( n \)
• B. 0
• C. \( -n \)
• D. 1

Hint:

Limits of the form \( \frac{f(x)-f(0)}{x} \) directly give the derivative at 0.

Answer 1

The Correct Option is C

Step 1: Recognize the limit form.
\[ \lim_{x\to 0} \frac{(1-x)^n - 1}{x} \] is of the form \( \frac{f(x)-f(0)}{x} \), which suggests derivative at \(x=0\).

Step 2: Define the function.
Let:
\[ f(x) = (1-x)^n. \]
Then:
\[ f(0) = 1. \]

Step 3: Use derivative definition.
\[ \lim_{x\to 0} \frac{f(x)-f(0)}{x} = f'(0). \]

Step 4: Differentiate the function.
\[ f'(x) = n(1-x)^{n-1}(-1). \]
\[ f'(x) = -n(1-x)^{n-1}. \]

Step 5: Evaluate at \(x=0\).
\[ f'(0) = -n(1)^{n-1}. \]
\[ f'(0) = -n. \]

Step 6: Substitute in limit.
Thus,
\[ \lim_{x\to 0} \frac{(1-x)^n - 1}{x} = -n. \]

Step 7: Final conclusion.
\[ \boxed{-n} \]