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

Question

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

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Answer 1 · 2018-01-25T10:58:57

See below.

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:

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

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

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

$lim_(xto+oo)[f(x)+f'(x)]$ $=λ$

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,

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