Question

\( \lim_{x\to0} \frac{1+x-e^x}{x^2} \)

• A. \( 1 \)
• B. \( -\frac{1}{2} \)
• C. \( 0 \)
• D. \( \frac{1}{2} \)
• E. \( -1 \)

Hint:

Use series expansion when direct substitution gives \(0/0\).

Answer 1

The Correct Option is B

Concept: Use Taylor expansion of \(e^x\).

Step 1: Expand: \[ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots \]

Step 2: Substitute: \[ 1+x - (1+x+\frac{x^2}{2}+...) \]

Step 3: Simplify: \[ = -\frac{x^2}{2} - \frac{x^3}{6} - ... \]

Step 4: Divide by \(x^2\): \[ = -\frac{1}{2} - \frac{x}{6} + ... \]

Step 5: Take limit: \[ x \to 0 \Rightarrow -\frac{1}{2} \]