A limited problem

Calculus Level 2

lim x 0 x ( 1 x + 2 x + 3 x + + 100 x ) = ? \lim_{x\to0} x \left( \left \lfloor \frac1x \right \rfloor + \left \lfloor \frac2x \right \rfloor + \left \lfloor \frac3x \right \rfloor + \cdots + \left \lfloor \frac{100}x \right \rfloor \right) = \, ?

Notation: \lfloor \cdot \rfloor denotes the floor function .


The answer is 5050.

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3 solutions

Chris Lewis
Aug 12, 2020

For any x > 0 x>0 and k > 0 k>0 , k x 1 < k x k x \frac{k}{x}-1<\left \lfloor \frac{k}{x} \right \rfloor \le \frac{k}{x}

Multiplying through by x x , k x < x k x k k-x<x\left \lfloor \frac{k}{x} \right \rfloor \le k

Now we can use the squeeze theorem; as x 0 x \to 0 , x k x k x\left \lfloor \frac{k}{x} \right \rfloor \to k .

Applying this to the sum in the question, the limit is 1 + 2 + + 100 = T 100 = 5050 1+2+\cdots+100=T_{100}=\boxed{5050}

Saúl Huerta
Aug 15, 2020

Note that:

a x = a x { a x } \left\lfloor\frac{a}{x}\right\rfloor=\frac{a}{x}-\left\{\frac{a}{x}\right\} Where { a x } \left\{\dfrac{a}{x}\right\} is the fractionary part of a x \dfrac{a}{x}

lim x 0 x [ ( 1 x + 2 x + . . . + 100 x ) ( { 1 x } + { 2 x } + . . . + { 100 x } ) ] \implies \lim_{x\rightarrow 0}x\left[\left(\frac{1}{x}+\frac{2}{x}+...+\frac{100}{x}\right)-\left(\left\{\frac{1}{x}\right\}+\left\{\frac{2}{x}\right\}+...+\left\{\frac{100}{x}\right\}\right)\right] lim x 0 ( 1 + 2 + 3 + . . . + 100 ) ( x { 1 x } + x { 2 x } + . . . + x { 100 x } ) \lim_{x\rightarrow 0}\left(1+2+3+...+100\right)-\left(x\left\{\frac{1}{x}\right\}+x\left\{\frac{2}{x}\right\}+...+x\left\{\frac{100}{x}\right\}\right)

Now, there are two cases. If ( a x ) \left(\dfrac{a}{x}\right) is an integer, then { a x } = 0 \left\{\dfrac{a}{x}\right\}=0 and consecuently x { a x } = 0 x\left\{\dfrac{a}{x}\right\}=0 .

If a x \dfrac{a}{x} is not an integer, then 0 < { a x } < 1 0<\left\{\dfrac{a}{x}\right\}<1 , 0 < x { a x } < x \implies0<x\left\{\dfrac{a}{x}\right\}<x

In the second case as x 0 x\rightarrow 0 x { a x } 0 \implies x\left\{\dfrac{a}{x}\right\}\rightarrow 0 . It gets squeezed to 0.

So all in all, the limit will just be the sum of the integers from 1 to 100 , as the other terms go to 0.

Nice method indeed.

Since x 0 x \rightarrow 0 , the denominator is infinitesimally small and hence the quotient approaches closer and closer to the numerator.

So it can be rewritten as 1 + 2 + 3 + . . . . + 100 1+ 2+ 3+ .... + 100

Which will be equal to 50 × 101 = 5050 50× 101 = 5050

Non fair solution.

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I know, I'm just a newbee, but exactly where does it go wrong, I post all of my solutions only to know this.

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Hello Devbrat,

The part which your solution lacks is rigour. While intuition does serve a purpose in leading your thought process to a certain direction, a clear proof with a mathematical groundwork is very often much necessary in formal proofs.

Sometimes intuitions could lead us astray such as the famous sum n = 1 1 n \sum_{n=1}^{\infty} \frac{1}{n} which intuitively seems to be convergent to some value as the terms themselves converge to zero, but the sum has been proved to diverge.

Take an example of the other solution to this question which has been posted by Chris Lewis. He uses the formal mathematical groundwork by using two things:

a) he uses the definition of a floor function to generate two inequalities within which our required expression is bounded, and

b) he uses the Squeeze theorem (aka Sandwich theorem) to prove that the limit is indeed the integer in the numerator for every term

while he doesn't provide the proof for this theorem, it is in fact a well-known theorem and the proofs can be referred to in some mathematical work focusing on this topic.

Hence, he uses more than simply intuition to drive his proof. In fact, it's even possible he never had an intuition like yours and just used the known mathematical theorems and definitions to come up with the proof.

To build solid proofs which support your deduction, you should read and practice more so as to get well-equipped with the right mathematical groundwork before tackling a problem. Once you start indulging in proof-writing you'll understand the difference between the two and why the kind of proofs like Chris has written above are given more importance than intuition.

Hope this helps.

Tapas Mazumdar - 10 months ago

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