Evaluating limit of floor function

[Raghav Gupta]

Homework Statement

How $$ \lim_{x\to\infty} \frac{nlogx}{[x]} = 0 $$ ? Here n∈ ℕ
Here [x] is greatest integer or floor function of x.

Homework Equations

[x] = x - {x} where {x} is fractional part of x.

The Attempt at a Solution

I know floor function is not differentiable.
We are getting here ∞/∞ form but can't apply L hopital rule as denominator is not differentiable.
How the limit evaluates to zero?

 

[AdityaDev]

use sandwich theorem.
hint: start from $?\le [x]\le ?$

 

[Raghav Gupta]

use sandwich theorem.
hint: start from $?\le [x]\le ?$

0≤ {x} < 1
Now what?

 

[AdityaDev]

No. i am asking about [x] and not {x}.

 

[Raghav Gupta]

Okay,
-∞≤[x] ≤ ∞ but [x] ≠ fractions

 

[AdityaDev]

No. Write it as $f(x)\le [x]\le g(x)$. what is f and g?

 

[Raghav Gupta]

No. Write it as $f(x)\le [x]\le g(x)$. what is f and g?

Why, what is f(x) and g(x) here?

 

[AdityaDev]

Okay. Use $x-1\le [x]\le x$

 

[Raghav Gupta]

Okay. Use $x-1\le [x]\le x$

Okay, what next?

 

[AdityaDev]

try to transform that inequality and make it look like $\frac{lnx}{[x]}$ by taking reciprocal and multiplying by lnx. will the inequality change?

 

[Raghav Gupta]

try to transform that inequality and make it look like $\frac{lnx}{[x]}$ by taking reciprocal and multiplying by lnx. will the inequality change?

Then
1/x≤ 1/[x]≤ 1/(x-1)
lnx/x ≤ lnx/[x]≤ lnx/(x-1)
Now?

 

[AdityaDev]

Good. Now from sandwich theorem, if you have $f(x)\le g(x)\le h(x)$ and $\lim_{x \to a}f(x)=L=\lim_{x \to a}h(x)$, then you can say that $\lim_{x \to a}g(x) = L$.

 

[Raghav Gupta]

Good. Now from sandwich theorem, if you have $f(x)\le g(x)\le h(x)$ and $\lim_{x \to a}f(x)=L=\lim_{x \to a}h(x)$, then you can say that $\lim_{x \to a}g(x) = L$.

Oh, great lnx/x will be zero when x approach infinity same with h(x). So, g(x) evaluates to zero. Thanks.

 

[Mark44]

Oh, great lnx/x will be zero when x approach infinity same with h(x). So, g(x) evaluates to zero. Thanks.

Let's be more precise. ln(x)/x never becomes zero, but gets arbitrarily close to zero as x grows large without bound.
Also, here's the proper notation:
$\lfloor x \rfloor$
\lfloor x \rfloor

For the ceiling function, it is:
$\lceil x \rceil$
\lceil x \rceil