Question

question 10 (6 points)
evaluate the limit:
\\\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}\\
enter an exact value.

Explanation:

Identify the limit form

Using the LHopitals Rule knowledge point

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

Direct substitution:

$$ \frac{e^0 - 1 - 0}{0^2} = \frac{1 - 1 - 0}{0} = \frac{0}{0} $$

Apply L'Hopital's Rule first time

Using the LHopitals Rule knowledge point

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

Direct substitution:

$$ \frac{e^0 - 1}{2(0)} = \frac{0}{0} $$

Apply L'Hopital's Rule second time

Using the LHopitals Rule knowledge point

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

Evaluate the final limit

Using the LHopitals Rule knowledge point

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

Answer:

Evaluate the limit:

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

Enter an exact value.
<blank>\(\frac{1}{2}\)</blank>