Question

What is the limit of `(2x+3)/(5x+7)` as x goes to infinity?

Answer 1

`lim(x->oo)(2x+3)/(5x+7)=2/5`

Explanation:

`lim(x->oo)(2x+3)/(5x+7)`

Notice that the degree of the numerator and the denominator are

the same i.e: 1, for this and all the similar scenarios the limits is

simply the ratio of the leading coefficients of top to bottom:

`:.lim(x->oo)(2x+3)/(5x+7)=2/5`

Answer 2

As `x->oo`, the `3` and `7` become insignificant relative to the magnitude of `x`. In other words, `x` becomes so large that:

`lim_(x->oo) (2x + 3)/(5x + 7) = lim_(x->oo) (2color(red)(cancel(color(black)(x))))/(5color(red)(cancel(color(black)(x))))`

`= lim_(x->oo) 2/5`

`= color(blue)(2/5)`