Question

How do you find the limit of `n/(3n-1)` as n approaches infinity?

Answer 1

`lim_(nrarroo) n/(3n-1) = 1/3`

Explanation:

Dividing the numerator and denominator of `n/(3n-1)` by `n` gives
`color(white)("XXX")n/(3n-1)=(n/n)/((3n)/n-1/n) = 1/(3-1/n)`

Therefore
`color(white)("XXX")lim_(nrarroo)n/(3n-1)`

`color(white)("XXX")=lim_(nrarroo)1/(3-1/n)`

`color(white)("XXX")=1/3`