Question

How do you evaluate `(2e^x+6)/(7e^x+5)` as x approaches infinity?

Answer 1

`lim_(xrarroo) (2e^x+6)/(7e^x+5) = 2/7`

Explanation:

`(2e^x+6)/(7e^x+5) = (2+6/(e^x))/(7+5/(e^x))`

As `xrarroo`
`color(white)("XXX")6/(e^x) rarr 0`
and
`color(white)("XXX")5/(e^x) rarr 0`

So
`color(white)("XXX")lim_(xrarroo) (2e^x+6)/(7e^x+5) = (2+0)/(7+0) = 2/7`