Questions: Evaluate the limit using L'Hôpital's rule [ lim x rightarrow 0 frace^x+x-19 x ]

Solution Steps

To evaluate the limit using L'Hôpital's rule, first check if the limit is in an indeterminate form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). If it is, differentiate the numerator and the denominator separately and then take the limit again. Repeat this process if necessary until the limit can be evaluated directly.

We start by evaluating the limit: \[ \lim_{x \rightarrow 0} \frac{e^{x} + x - 1}{9x} \] Substituting \( x = 0 \) gives us: \[ \frac{e^{0} + 0 - 1}{9 \cdot 0} = \frac{1 + 0 - 1}{0} = \frac{0}{0} \] This is an indeterminate form, so we can apply L'Hôpital's rule.

We differentiate the numerator and the denominator:

• The derivative of the numerator \( e^{x} + x - 1 \) is \( e^{x} + 1 \).
• The derivative of the denominator \( 9x \) is \( 9 \).

Now we apply L'Hôpital's rule: \[ \lim_{x \rightarrow 0} \frac{e^{x} + 1}{9} \] Substituting \( x = 0 \) gives us: \[ \frac{e^{0} + 1}{9} = \frac{1 + 1}{9} = \frac{2}{9} \]

Final Answer