Question

Show that `lim_(x to +oo)f'(x)=0` ?

Given `f` differentiable in `RR` with `f(x)!=0` `AAx` `in` `RR`. If `lim_(x to +oo)f(x)` exists then show that `lim_(x to +oo)f'(x)=0`

Answer 1

See below.

Explanation:

Solved it.

`lim_(xto+oo)f(x)` `in` `RR`

Supposed `lim_(xto+oo)f(x)=λ`

then `lim_(xto+oo)f(x)=lim_(xto+oo)(e^xf(x))/e^x`

We have `((+-oo)/(+oo))` and `f` is differentiable in `RR` so applying Rules De L'Hospital:

`lim_(xto+oo)(e^xf(x))/e^x=`

`lim_(xto+oo)(e^xf(x)+e^xf'(x))/e^x=`

`lim_(xto+oo)((e^xf(x))/e^x+(e^xf'(x))/e^x)=`

`lim_(xto+oo)[f(x)+f'(x)]` `=λ`

• `h(x)=f(x)+f'(x)` with `lim_(xto+oo)h(x)=λ`

Thus, `f'(x)=h(x)-f(x)`

Therefore, `lim_(xto+oo)f'(x)=lim_(xto+oo)[h(x)-f(x)]`

`=λ-λ=0`

As a result,

`lim_(xto+oo)f'(x)=0`