Limits, Continuity and Differentiability Part 2

As, we know that,
r≤[r]<r+1
2r≤[2r]<2r+1
3r≤[3r]<3r+1
:   :  :

nr≤[nr]<nr+1
Adding; Now,

lim

n→∞

‌

n(n+1)⋅r

2⋅n2

=

lim

n→∞

‌

n⋅n(1+ 1 n )⋅r

2n2

=‌

(1+0)⋅r

2

=‌

r

2

....(i)
and =‌

(1+0)⋅r+0

2

=‌

r

2

...(ii)
From Eqs. (i) and (ii), by Sandwich theorem, we conclude that,
⇒‌‌

lim

n→∞

‌

[r]+[2r]+[3r]+...+[nr]

n2

=‌

r

2

Sandwich Theorem ⇒ Let g(x)≤f(x)≤h(x)
and

lim

x→a

g(x)=

lim

x→a

h(x)=1
∴‌‌

lim

x→a

f(x)=1