Evaluate the limit using L'Hôpital's rule $$ \lim _{x \rightarrow 0} \frac{e^{x}+x-1}{9 x} $$

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1. First, identify that as $$x \rightarrow 0$$:
   - The numerator becomes $$e^0 + 0 - 1 = 1 + 0 - 1 = 0$$.
   - The denominator becomes $$9(0) = 0$$.

2. Since the limit is in the indeterminate form $$\frac{0}{0}$$, we apply L'Hôpital's rule, which involves differentiating the numerator and the denominator.

3. Differentiate the numerator and denominator with respect to $$x$$:
   - Numerator derivative: 
     $$
     \frac{d}{dx}(e^x + x - 1) = e^x + 1
     $$
   - Denominator derivative:
     $$
     \frac{d}{dx}(9x) = 9
     $$

4. Apply L'Hôpital's rule to obtain a new limit:
   $$
   \lim_{x \rightarrow 0} \frac{e^x + 1}{9}
   $$

5. Substitute $$x = 0$$ into the differentiated expression:
   $$
   \frac{e^0 + 1}{9} = \frac{1 + 1}{9} = \frac{2}{9}
   $$

6. Therefore, the evaluated limit is:
   $$
   \boxed{\frac{2}{9}}
   $$
The limit is $$\frac{2}{9}$$.

Evaluate the limit using L’Hôpital’s rule

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