Evaluate the following limits :
`Lim_( x to 1) x^(1/(x-1))`

Answer

To evaluate the limit \( \lim_{x \to 1} x^{\frac{1}{x-1}} \), we can follow these steps:

### Step 1: Identify the Form of the Limit
First, we substitute \( x = 1 \) into the expression:
\[ x^{\frac{1}{x-1}} = 1^{\frac{1}{1-1}} = 1^{\infty} \]
This is an indeterminate form.

### Step 2: Take the Natural Logarithm
Let \( L = \lim_{x \to 1} x^{\frac{1}{x-1}} \). Taking the natural logarithm on both sides gives:
\[ \log L = \lim_{x \to 1} \frac{\log x}{x - 1} \]

### Step 3: Evaluate the Limit
Now we need to evaluate \( \lim_{x \to 1} \frac{\log x}{x - 1} \). Substituting \( x = 1 \) gives \( \frac{0}{0} \), which is another indeterminate form. We can apply L'Hôpital's Rule.

### Step 4: Apply L'Hôpital's Rule
Using L'Hôpital's Rule, we differentiate the numerator and the denominator:
\[ \log x \rightarrow \frac{1}{x} \quad \text{and} \quad x - 1 \rightarrow 1 \]
So we have:
\[ \log L = \lim_{x \to 1} \frac{\frac{1}{x}}{1} = \lim_{x \to 1} \frac{1}{x} = 1 \]

### Step 5: Solve for \( L \)
Now we can exponentiate both sides to solve for \( L \):
\[ L = e^{\log L} = e^1 = e \]

### Final Result
Thus, the limit is:
\[ \lim_{x \to 1} x^{\frac{1}{x-1}} = e \]

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