Question

Can anyone help me factor $lim\\\\x--\ \textgreater \ 1\\\\\frac{x^4-1}{x-1}$

as well as: →$lim\\x->1 \\\\\\\frac{x-1}{x^3-x^2+x-1}$

Answer 1

For the first limit, you can factor the numerator as a difference of squares:

$x^4-1=(x^2-1)(x^2+1)$

and the first factor can be factored the same way:

$x^4-1=(x-1)(x+1)(x^2+1)$

More generally, we have the formula

$1+x+x^2+\cdots+x^n=\dfrac{x^{n+1}-1}{x-1}$

so really, for $x\neq1$,

$\dfrac{x^4-1}{x-1}=x^3+x^2+x+1$

In any case, we get

$\displaystyle\lim_{x\to1}\frac{x^4-1}{x-1}=\lim_{x\to1}(x+1)(x^2+1)=4$

For the second limit, you can factor the denominator by grouping:

$x^3-x^2+x-1=(x^3-x^2)+(x-1)=x^2(x-1)+(x-1)=(x^2+1)(x-1)$

Then the limit is

$\displaystyle\lim_{x\to1}\frac{x-1}{x^3-x^2+x-1}=\lim_{x\to1}\frac1{x^2+1}=\frac12$