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Factor the difference of squares $x^2-1$ as the product of two conjugated binomials
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$1=\lim_{x\to1}\left(\frac{\left(x+1\right)\left(x-1\right)}{x^3-1}=\frac{2}{3}\right)$
Learn how to solve problems step by step online. Find the limit 1=(x)->(1)lim((x^2-1)/(x^3-1)=2/3). Factor the difference of squares x^2-1 as the product of two conjugated binomials. Factor the difference of cubes: a^3-b^3 = (a-b)(a^2+ab+b^2). Simplify the fraction . Evaluate the limit \lim_{x\to1}\left(\frac{x+1}{x^2+x+1}=\frac{2}{3}\right) by replacing all occurrences of x by 1.