Evaluate the following limits :
`Lim_(xto 1) (x^(m)-1)/(x^(n)-1)`

Answer

To evaluate the limit

\[ \lim_{x \to 1} \frac{x^m - 1}{x^n - 1}, \]

we first observe that substituting \(x = 1\) directly into the expression gives us:

\[ \frac{1^m - 1}{1^n - 1} = \frac{0}{0}, \]

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 \(f(x) = x^m - 1\) and the denominator is \(g(x) = x^n - 1\).

Differentiating both:

- The derivative of the numerator \(f'(x) = \frac{d}{dx}(x^m - 1) = mx^{m-1}\).
- The derivative of the denominator \(g'(x) = \frac{d}{dx}(x^n - 1) = nx^{n-1}\).

### Step 2: Apply L'Hôpital's Rule

Now we can apply L'Hôpital's Rule:

\[ \lim_{x \to 1} \frac{x^m - 1}{x^n - 1} = \lim_{x \to 1} \frac{f'(x)}{g'(x)} = \lim_{x \to 1} \frac{mx^{m-1}}{nx^{n-1}}. \]

### Step 3: Substitute \(x = 1\)

Now we substitute \(x = 1\):

\[ \lim_{x \to 1} \frac{mx^{m-1}}{nx^{n-1}} = \frac{m \cdot 1^{m-1}}{n \cdot 1^{n-1}} = \frac{m}{n}. \]

### Final Result

Thus, the limit evaluates to:

\[ \lim_{x \to 1} \frac{x^m - 1}{x^n - 1} = \frac{m}{n}. \]