Menu Close

is-lim-x-a-f-x-lim-x-a-f-x-




Question Number 89576 by M±th+et£s last updated on 18/Apr/20
is  lim_(x→a) ⌊f(x)⌋=⌊lim_(x→a )  f(x)⌋
$${is}\:\:\underset{{x}\rightarrow{a}} {{lim}}\lfloor{f}\left({x}\right)\rfloor=\lfloor\underset{{x}\rightarrow{a}\:} {{lim}}\:{f}\left({x}\right)\rfloor\: \\ $$
Commented by mr W last updated on 18/Apr/20
i think no.  example f(x)=((sin x)/x)  lim_(x→0) ⌊f(x)⌋=0, since 0<((sin x)/x)<1 ⇒⌊((sin x)/x)⌋=0  ⌊lim_(x→0) f(x)⌋=1, since lim_(x→0) (((sin x)/x))=1
$${i}\:{think}\:{no}. \\ $$$${example}\:{f}\left({x}\right)=\frac{\mathrm{sin}\:{x}}{{x}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\lfloor{f}\left({x}\right)\rfloor=\mathrm{0},\:{since}\:\mathrm{0}<\frac{\mathrm{sin}\:{x}}{{x}}<\mathrm{1}\:\Rightarrow\lfloor\frac{\mathrm{sin}\:{x}}{{x}}\rfloor=\mathrm{0} \\ $$$$\lfloor\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{f}\left({x}\right)\rfloor=\mathrm{1},\:{since}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{sin}\:{x}}{{x}}\right)=\mathrm{1} \\ $$
Commented by M±th+et£s last updated on 18/Apr/20
yes sir it was typo and thank you
$${yes}\:{sir}\:{it}\:{was}\:{typo}\:{and}\:{thank}\:{you} \\ $$
Commented by mr W last updated on 18/Apr/20
nice to see that you can edit your post.
$${nice}\:{to}\:{see}\:{that}\:{you}\:{can}\:{edit}\:{your}\:{post}. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *