Evaluate `lim_(xto a)(log{1+(x-a)})/((x-a))`

Answer

To evaluate the limit

\[ \lim_{x \to a} \frac{\log(1 + (x - a))}{(x - a)}, \]

we start by substituting \( x = a \) into the expression. This gives us:

\[ \frac{\log(1 + (a - a))}{(a - a)} = \frac{\log(1)}{0} = \frac{0}{0}, \]

which is an indeterminate form. Therefore, we can apply L'Hôpital's Rule, which states that if we have a limit of the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can differentiate the numerator and denominator separately.

### Step 1: Differentiate the numerator and denominator

The numerator is \( \log(1 + (x - a)) \) and the denominator is \( (x - a) \).

1. Differentiate the numerator:
\[ \frac{d}{dx} \left( \log(1 + (x - a)) \right) = \frac{1}{1 + (x - a)} \cdot \frac{d}{dx}(x - a) = \frac{1}{1 + (x - a)}. \]

2. Differentiate the denominator:
\[ \frac{d}{dx}(x - a) = 1. \]

### Step 2: Apply L'Hôpital's Rule

Now we can apply L'Hôpital's Rule:

\[ \lim_{x \to a} \frac{\log(1 + (x - a))}{(x - a)} = \lim_{x \to a} \frac{\frac{1}{1 + (x - a)}}{1}. \]

### Step 3: Evaluate the limit

Now we substitute \( x = a \):

\[ \lim_{x \to a} \frac{1}{1 + (x - a)} = \frac{1}{1 + (a - a)} = \frac{1}{1} = 1. \]

### Final Answer

Thus, the limit is

\[ \boxed{1}. \]

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