Can anyone help me factor [tex]lim\\\\x--\ \textgreater \ 1\\\\\frac{x^4-1}{x-1}[/tex]as well as: →[tex]lim\\x->1 \\\\\\\frac{x-1}{x^3-x^2+x-1}[/tex]

For the first limit, you can factor the numerator as a difference of squares:[tex]x^4-1=(x^2-1)(x^2+1)[/tex]and the first factor can be factored the same way:[tex]x^4-1=(x-1)(x+1)(x^2+1)[/tex]More generally, we have the formula[tex]1+x+x^2+\cdots+x^n=\dfrac{x^{n+1}-1}{x-1}[/tex]so really, for [tex]x\neq1[/tex],[tex]\dfrac{x^4-1}{x-1}=x^3+x^2+x+1[/tex]In any case, we get[tex]\displaystyle\lim_{x\to1}\frac{x^4-1}{x-1}=\lim_{x\to1}(x+1)(x^2+1)=4[/tex]For the second limit, you can factor the denominator by grouping:[tex]x^3-x^2+x-1=(x^3-x^2)+(x-1)=x^2(x-1)+(x-1)=(x^2+1)(x-1)[/tex]Then the limit is[tex]\displaystyle\lim_{x\to1}\frac{x-1}{x^3-x^2+x-1}=\lim_{x\to1}\frac1{x^2+1}=\frac12[/tex]