Question

1. $\lim_{x\to1}\frac{x^{3}-1}{x^{2}-1}$

Explanation:

Step1: Factorize numerator and denominator

Use the formulas \(a^{3}-b^{3}=(a - b)(a^{2}+ab + b^{2})\) and \(a^{2}-b^{2}=(a - b)(a + b)\).
For \(x^{3}-1\), \(a=x\), \(b = 1\), so \(x^{3}-1=(x - 1)(x^{2}+x + 1)\).
For \(x^{2}-1\), \(a=x\), \(b = 1\), so \(x^{2}-1=(x - 1)(x + 1)\).
Then \(\frac{x^{3}-1}{x^{2}-1}=\frac{(x - 1)(x^{2}+x + 1)}{(x - 1)(x + 1)}\).
Since \(x\to1\) but \(x
eq1\), we can cancel out the non - zero factor \((x - 1)\).

Step2: Substitute \(x = 1\) into the simplified function

After canceling \((x - 1)\), the function becomes \(\frac{x^{2}+x + 1}{x + 1}\).
Substitute \(x = 1\) into \(\frac{x^{2}+x + 1}{x + 1}\), we get \(\frac{1^{2}+1 + 1}{1+1}=\frac{3}{2}\).

Answer:

\(\frac{3}{2}\)