`lim_(x to 0)(ln(1+x))/(x^(2))+(x-1)/(x)=`

Options

A. `oo`
B. `(1)/(2)`
C. `-(1)/(2)`
D. 1

Answer

To solve the limit \( \lim_{x \to 0} \left( \frac{\ln(1+x)}{x^2} + \frac{x-1}{x} \right) \), we will break it down into manageable steps.

### Step 1: Simplify the expression
We can rewrite the limit as:
\[ \lim_{x \to 0} \left( \frac{\ln(1+x)}{x^2} + \frac{x-1}{x} \right) = \lim_{x \to 0} \left( \frac{\ln(1+x)}{x^2} + 1 - \frac{1}{x} \right) \]

### Step 2: Use the Taylor series expansion for \( \ln(1+x) \)
The Taylor series expansion of \( \ln(1+x) \) around \( x=0 \) is:
\[ \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \]
Substituting this into our limit gives:
\[ \lim_{x \to 0} \left( \frac{x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots}{x^2} + 1 - \frac{1}{x} \right) \]

### Step 3: Simplify the fraction
Now, we can simplify the fraction:
\[ \frac{x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots}{x^2} = \frac{x}{x^2} - \frac{\frac{x^2}{2}}{x^2} + \frac{\frac{x^3}{3}}{x^2} - \frac{\frac{x^4}{4}}{x^2} + \cdots \]
This simplifies to:
\[ \frac{1}{x} - \frac{1}{2} + \frac{x}{3} - \frac{x^2}{4} + \cdots \]

### Step 4: Combine the terms
Now, we substitute this back into our limit:
\[ \lim_{x \to 0} \left( \left( \frac{1}{x} - \frac{1}{2} + \frac{x}{3} - \frac{x^2}{4} + \cdots \right) + 1 - \frac{1}{x} \right) \]
The \( \frac{1}{x} \) terms cancel out:
\[ \lim_{x \to 0} \left( -\frac{1}{2} + 1 + \frac{x}{3} - \frac{x^2}{4} + \cdots \right) \]

### Step 5: Evaluate the limit
As \( x \to 0 \), the terms involving \( x \) vanish:
\[ -\frac{1}{2} + 1 = \frac{1}{2} \]

### Final Answer
Thus, the limit is:
\[ \boxed{\frac{1}{2}} \]