Question

Question #acaf1

Answer 1

`lim_(xrarr1) (x^4-1)/(x^3)=0`

Explanation:

Maybe I've misinterpretted the question
but `(x^4-1)/(x^3)` is defined when `x=1`
so
`color(white)("XXX")lim_(xrarr1)(x^4-1)/(x^3)=(1^4-1)/(1^3)=(1-1)/1=0/1=0`

Answer 2

`lim_(xrarr1)(4x-1)/(3x)=1`

Explanation:

This question seems to have been changed since I initially answered it (ignore my answer, probably below this one).

`(4x-1)/(3x)` is defined for `x=1`

So `lim_(xrarr1)(4x-1)/(3x)` is simply `(4x-1)/(3x)` with `x=1`

`color(white)("XXX")(4 * (1)-1)/(3 * (1))=3/3=1`