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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 *

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 *

Grade:12th pass

1 Answers

Samyak Jain
333 Points
2 years ago
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, limx\rightarrow0 x([1/x] + [2/x] + … + [k/x]) can be written as limx\rightarrow\infty ([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 limx\rightarrow\infty {k(k + 1)/2}/x  –  k  <  limx\rightarrow\infty ([1x] + [2x] + … + [kx])  \leq  limx\rightarrow\infty {k(k + 1)/2}/x
As limx\rightarrow\infty {k(k + 1)/2}/x  –  k  =  limx\rightarrow\infty {k(k + 1)/2}/x  =  k(k+1)/2,
by sandwich theorem, limx\rightarrow\infty ([1x] + [2x] + … + [kx]) = k(k+1)/2
i.e.  limx\rightarrow0 x([1/x] + [2/x] + … + [k/x])  =  k(k+1)/2

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