Evaluate the following limits :
`lim_(x to 1)(x^(15)-1)/(x^(10)-1)`

Answer

To evaluate the limit

\[ \lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1}, \]

we will follow these steps:

### Step 1: Direct Substitution
First, we substitute \( x = 1 \) into the limit:

\[ \frac{1^{15} - 1}{1^{10} - 1} = \frac{1 - 1}{1 - 1} = \frac{0}{0}. \]

This gives us an indeterminate form \( \frac{0}{0} \), which means we cannot evaluate the limit directly.

**Hint:** If you get a \( \frac{0}{0} \) form, consider using L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if \( \lim_{x \to c} \frac{f(x)}{g(x)} \) results in \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then:

\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}, \]

provided the limit on the right side exists.

### Step 3: Differentiate the Numerator and Denominator
Now, we differentiate the numerator and the denominator:

- The derivative of the numerator \( f(x) = x^{15} - 1 \) is:

\[ f'(x) = 15x^{14}. \]

- The derivative of the denominator \( g(x) = x^{10} - 1 \) is:

\[ g'(x) = 10x^{9}. \]

### Step 4: Substitute Again
Now we substitute back into the limit:

\[ \lim_{x \to 1} \frac{f'(x)}{g'(x)} = \lim_{x \to 1} \frac{15x^{14}}{10x^{9}}. \]

Substituting \( x = 1 \):

\[ \frac{15 \cdot 1^{14}}{10 \cdot 1^{9}} = \frac{15}{10} = \frac{3}{2}. \]

### Conclusion
Thus, the limit evaluates to:

\[ \lim_{x \to 1} \frac{x^{15} - 1}{x^{10} - 1} = \frac{3}{2}. \]

**Final Answer:** \( \frac{3}{2} \)

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