Let's solve the limit step-by-step using **L'Hôpital's rule**: Given: \[ \lim_{x \to 0} \frac{e^x + x - 1}{9x} \] ### Step 1: Check the Indeterminate Form Substitute \(x = 0\): Numerator: \(e^0 + 0 - 1 = 1 + 0 - 1 = 0\) Denominator: \(9 \times 0 = 0\) So, the limit is of the indeterminate form \(\frac{0}{0}\), so we **can use L'Hôpital's rule**. --- ### Step 2: Differentiate Numerator and Denominator Differentiate numerator: \[ \frac{d}{dx}(e^x + x - 1) = e^x + 1 \] Differentiate denominator: \[ \frac{d}{dx}(9x) = 9 \] --- ### Step 3: Apply L'Hôpital's Rule \[ \lim_{x \to 0} \frac{e^x + x - 1}{9x} = \lim_{x \to 0} \frac{e^x + 1}{9} \] --- ### Step 4: Substitute \(x = 0\) \[ \frac{e^0 + 1}{9} = \frac{1 + 1}{9} = \frac{2}{9} \] --- ### **Final Answer** \[ \boxed{\frac{2}{9}} \]

Let's carefully evaluate the limit step-by-step using **L'Hôpital's rule**. --- ### **Given limit:** \[ \lim_{x \to 0} \frac{e^x + x - 1}{9x} \] --- ### **Step 1: Check the indeterminate form** Substitute \(x = 0\): - Numerator: \(e^0 + 0 - 1 = 1 + 0 - 1 = 0\) - Denominator: \(9 \times 0 = 0\) Since the limit yields \(\frac{0}{0}\), an indeterminate form, **L'Hôpital's rule** can be applied. --- ### **Step 2: Differentiate numerator and denominator** - Derivative of numerator: \[ \frac{d}{dx}(e^x + x - 1) = e^x + 1 \] - Derivative of denominator: \[ \frac{d}{dx}(9x) = 9 \] --- ### **Step 3: Rewrite the limit after differentiation** Applying L'Hôpital's rule: \[ \lim_{x \to 0} \frac{e^x + 1}{9} \] Now, directly substitute \(x = 0\): \[ \frac{e^0 + 1}{9} = \frac{1 + 1}{9} = \frac{2}{9} \] --- ### **Final answer:** \[ \boxed{\frac{2}{9}} \] **Thus, the value of the limit is \(\boxed{\frac{2}{9}}\).**