Question

Can anyone help me factor
`lim\\\\x--\ \textgreater \ 1\\\\(x^4-1)/(x-1)`

as well as: →
`lim\\x->1 \\\\\\(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=(x^(n+1)-1)/(x-1)`

so really, for
`x\\eq1`,

`(x^4-1)/(x-1)=x^3+x^2+x+1`

In any case, we get

`\displaystyle\lim_(x\to1)(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)(x-1)/(x^3-x^2+x-1)=\lim_(x\to1)\frac1{x^2+1}=\frac12`