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Factor the difference of squares $x^4-1$ as the product of two conjugated binomials
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$\lim_{x\to1}\left(\frac{\left(x^{2}+1\right)\left(x^{2}-1\right)}{x^6-1}\right)$
Learn how to solve limits by direct substitution problems step by step online. Find the limit of (x^4-1)/(x^6-1) as x approaches 1. Factor the difference of squares x^4-1 as the product of two conjugated binomials. Factor the difference of squares \left(x^{2}-1\right) as the product of two conjugated binomials. Factor the sum or difference of cubes using the formula: a^3\pm b^3 = (a\pm b)(a^2\mp ab+b^2). Simplify \frac{\left(x^{2}+1\right)\left(x+1\right)\left(x-1\right)}{\left(x^{2}-1\right)\left(x^{4}+x^{2}+1\right)}.