Question

evaluate the limit using lhôpitals rule
\\\lim_{x \to 0} \frac{e^x + x - 1}{2x}\\

Explanation:

Verify the indeterminate form

$$ \lim_{x \to 0} (e^x + x - 1) = e^0 + 0 - 1 = 0 $$

$$ \lim_{x \to 0} (2x) = 0 $$

$$ \text{Form: } \frac{0}{0} $$

Apply L'Hôpital's Rule

$$ \lim_{x \to 0} \frac{\frac{d}{dx}(e^x + x - 1)}{\frac{d}{dx}(2x)} = \lim_{x \to 0} \frac{e^x + 1}{2} $$

Evaluate the limit

$$ \lim_{x \to 0} \frac{e^x + 1}{2} = \frac{e^0 + 1}{2} = \frac{1 + 1}{2} = 1 $$

Answer:

Evaluate the limit using L'Hôpital's rule

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

<blank>1</blank>