Question

Lim x-> infinity 7/((e^x)+2) evaluate ?

Answer 1

`lim_(x->oo)7/(e^x+2)=0`

Explanation:

Let try to substitute `oo` for `x`

Recall: `lim_(x->oo)e^x=oo`

This can easily be verified by taking a look of the graph of `e^x` (See Below)

graph{e^x [-9.58, 10.42, -1.24, 8.76]}

`lim_(x->oo)7/(e^x+2)=7/(e^color(red)oo+2)=7/(oo+2)=7/oo=0`

As to why `7/oo` tends to `0` think of it like this:

As `x` approaches infinity the denominator gets larger and larger towards infinity but the fraction as a whole gets smaller and smaller. Here's a few examples:

Consider:

`1/10=0.1`

`1/100=0.01`

`1/1000=0.001`

`1/10000000=0.0000001`

As you can see, the fraction becomes smaller and smaller/ closer and closer to `0`. The fact that the numerator is `7` instead of `1` doesn't change this fact.