# Evaluating limit of floor function

1. May 14, 2015

### Raghav Gupta

1. The problem statement, all variables and given/known data

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

3. 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?

2. May 14, 2015

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

3. May 14, 2015

### Raghav Gupta

0≤ {x} < 1
Now what?

4. May 14, 2015

5. May 14, 2015

### Raghav Gupta

Okay,
-∞≤[x] ≤ ∞ but [x] ≠ fractions

6. May 14, 2015

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

7. May 14, 2015

### Raghav Gupta

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

8. May 14, 2015

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

9. May 14, 2015

### Raghav Gupta

Okay, what next?

10. May 14, 2015

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

11. May 14, 2015

### Raghav Gupta

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

12. May 14, 2015

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 $\lim_{x \to a}g(x) = L$.

13. May 14, 2015

### Raghav Gupta

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

14. May 14, 2015

### Staff: Mentor

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