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

Evaluate the limit using L'Hôpital's rule
[
lim x rightarrow 0 frace^x+x-19 x
]
Transcript text: Evaluate the limit using L'Hôpital's rule \[ \lim _{x \rightarrow 0} \frac{e^{x}+x-1}{9 x} \] $\square$
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Solution

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Solution Steps

To evaluate the limit using L'Hôpital's rule, first check if the limit is in an indeterminate form like 00 \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.

Step 1: Evaluate the Initial Limit

We start by evaluating the limit: limx0ex+x19x \lim_{x \rightarrow 0} \frac{e^{x} + x - 1}{9x} Substituting x=0 x = 0 gives us: e0+0190=1+010=00 \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.

Step 2: Differentiate the Numerator and Denominator

We differentiate the numerator and the denominator:

  • The derivative of the numerator ex+x1 e^{x} + x - 1 is ex+1 e^{x} + 1 .
  • The derivative of the denominator 9x 9x is 9 9 .
Step 3: Apply L'Hôpital's Rule

Now we apply L'Hôpital's rule: limx0ex+19 \lim_{x \rightarrow 0} \frac{e^{x} + 1}{9} Substituting x=0 x = 0 gives us: e0+19=1+19=29 \frac{e^{0} + 1}{9} = \frac{1 + 1}{9} = \frac{2}{9}

Final Answer

The limit is 29 \boxed{\frac{2}{9}}

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