100. $\\lim_{x\\to1}\\frac{x^{3}-1}{x^{2}-1}$
100. $\\lim_{x\\to1}\\frac{x^{3}-1}{x^{2}-1}$
Answer
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\neq1), 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})