(x^n-1)/((x-1) )=(x-α^1 )×(1-α^2 )×………..(x-α^((n-1)) )
How do we prove that n = the multiplication of the second term using limits and l'hopital's rule.
To prove that \(n\) is equal to the product of the second term, \(1 - \alpha^2\), we can use limits and L'Hôpital's rule.
Let's start by simplifying the expression on the left-hand side of the equation:
\[\frac{x^n-1}{x-1} = (x-α^1) \times (1-α^2) \times … \times (x-α^{n-1})\]
First, notice that when \(x = 1\), both sides of the equation are undefined due to division by zero. To avoid this issue, we will analyze the limit of the expression as \(x\) approaches 1.
Taking the limit as \(x\) approaches 1 on both sides of the equation, we have:
\[\lim_{x \to 1} \frac{x^n-1}{x-1} = \lim_{x \to 1} (x-α^1) \times (1-α^2) \times … \times (x-α^{n-1})\]
Now, let's apply L'Hôpital's rule to the left-hand side of the equation. Differentiate the numerator and the denominator with respect to \(x\):
\[\lim_{x \to 1} \frac{n \cdot x^{n-1}}{1} = \lim_{x \to 1} (x-α^1) \times (1-α^2) \times … \times (x-α^{n-1})\]
Since the denominator of the left-hand side is always equal to 1, we can simplify the equation further:
\[n = \lim_{x \to 1} (x-α^1) \times (1-α^2) \times … \times (x-α^{n-1})\]
Now, as \(x\) approaches 1, each term \(x-α^i\) on the right-hand side approaches \(1 - α^i\) since \(1-α\) is equal to \(0\). Therefore, the right-hand side becomes:
\[n = (1-α^1) \times (1-α^2) \times … \times (1-α^{n-1})\]
We have now proved that \(n\) is equal to the product of the second term, \(1-α^2\), using limits and L'Hôpital's rule.