lim x-0 x([1/x]+[2/x]...+[k/x]) by using sandwic

Question

lim x->0 x([1/x]+[2/x]...+[k/x]) by using sandwich theorem, i am getting k/2 on left and k^2/2 on right. can someone please help me with this. thank you so much *

Samyak Jain · Accepted Answer

According to me, both mentioned values are wrong.
Greatest integer is always less than or equal to the number.
x – 1 $<$ [x] $\leq$ x

Now, lim_x0 x([1/x] + [2/x] + … + [k/x]) can be written as lim_x ([1x] + [2x] + … + [kx]) / x

(x –1) + (2x –1) + … + (kx –1) $<$ ([1x] + [2x] + … + [kx]) $\leq$ x + 2x + … + kx

$\Rightarrow$ {k(k + 1)/2}/x – k $<$ ([1x] + [2x] + … + [kx]) $\leq$ {k(k + 1)/2}/x

Divide throughout by x and take limit as x $\rightarrow$ $\infty$ .

$\Rightarrow$ lim_x {k(k + 1)/2}/x – k $<$ lim_x ([1x] + [2x] + … + [kx]) $\leq$ lim_x {k(k + 1)/2}/x

As lim_x {k(k + 1)/2}/x – k = lim_x {k(k + 1)/2}/x = k(k+1)/2,

by sandwich theorem, lim_x ([1x] + [2x] + … + [kx]) = k(k+1)/2

i.e. lim_x0 x([1/x] + [2/x] + … + [k/x]) = k(k+1)/2

Integral Calculus> lim x->0 x([1/x]+[2/x]...+[k/x]) by using...

lim x->0 x([1/x]+[2/x]...+[k/x])

by using sandwich theorem, i am getting k/2 on left and k^2/2 on right. can someone please help me with this.

thank you so much *

shiya , 7 Years ago

Samyak Jain

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Integral Calculus> lim x->0 x([1/x]+[2/x]...+[k/x]) by using...

lim x->0 x([1/x]+[2/x]...+[k/x]) by using sandwich theorem, i am getting k/2 on left and k^2/2 on right. can someone please help me with this. thank you so much *

shiya , 7 Years ago

Samyak Jain

LIVE ONLINE CLASSES

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