Question

evaluate the limit using lhospitals rule
\\\lim_{x \to 0} \frac{e^x + x - 1}{7x}\\

Explanation:

Verify the indeterminate form

Using the LHopitals Rule knowledge point

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

$$ \lim_{x\to 0} (7x) = 7(0) = 0 $$

Since the limit yields the indeterminate form \(\frac{0}{0}\), L'Hospital's rule is applicable.

Differentiate numerator and denominator

Using the LHopitals Rule knowledge point

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

$$ \frac{d}{dx}(7x) = 7 $$

Evaluate the limit of the derivatives

Using the LHopitals Rule knowledge point

$$ \lim_{x\to 0} \frac{e^x + 1}{7} = \frac{e^0 + 1}{7} = \frac{1 + 1}{7} = \frac{2}{7} $$

Answer:

Evaluate the limit using L'Hospital's rule
\(\lim_{x\to 0} \frac{e^x + x - 1}{7x} =\) <blank>\(\frac{2}{7}\)</blank>