Question

What is `lim_(x->oo) (e^x-1)/x` ?

Answer 1

`lim_(x->oo) (e^x-1)/x = oo`

Explanation:

The Maclaurin expansion of `e^x = 1+x+x^2/(2!)+x^3/(3!)+ .......`

Hence, `e^x-1 = x+x^2/(2!)+x^3/(3!)+ .......`

`:. lim_(x->oo) (e^x-1)/x = lim_(x->oo) ((x+x^2/(2!)+x^3/(3!)+ ......)/x)`

`=lim_(x->oo) (1 + x/(2!) +(x^2)/(3!) + .......)`

`= oo`

Answer 2

`lim_(x->oo) (e^x-1)/x=oo`

Explanation:

If we consider the numerator and denominator we see that `e^x-1` will grow much much faster than `x` when `x` is large.

This means that the numerator will "outrun" the denominator and the gap will be getting larger and larger, so at infinity, the denominator will just be insignificant, leaving us with:
`lim_(x->oo) (e^x-1)/x=lim_(x->oo) e^x-1=oo`