MHB Limit of function containing integer part

[mathmari]

Hey!

I want to calculate the limit $$\lim_{x\rightarrow \infty}x^{100}\left [\frac{1}{x}\right ]$$

When $x\rightarrow +\infty$ it holds that $0<\frac{1}{x}<1$, or not? (Wondering)

If yes, it holds that $\left [\frac{1}{x}\right ]=0$ or not? Then $x^{100}\left [\frac{1}{x}\right ]=0$, and therefore the limit is $0$. But what happens if $x\rightarrow -\infty$ ? (Wondering)

 

[Euge]

When $x\rightarrow +\infty$ it holds that $0<\frac{1}{x}<1$, or not? (Wondering)

If you meant to say $0 < \frac{1}{x} < 1$ when $x$ is sufficiently large, then you are correct.

If yes, it holds that $\left [\frac{1}{x}\right ]=0$ or not? Then $x^{100}\left [\frac{1}{x}\right ]=0$, and therefore the limit is $0$.

That's correct.

But what happens if $x\rightarrow -\infty$ ? (Wondering)

When $x \le -1$, $-1 \le 1/x < 0$, which implies $\left[\frac{1}{x}\right] = -1$. Thus $x^{100}\left[\frac{1}{x}\right] = -x^{100}$ whenever $x \le -1$. What does that tell you about the limit as $x \to -\infty$?

 

[mathmari]

When $x \le -1$, $-1 \le 1/x < 0$, which implies $\left[\frac{1}{x}\right] = -1$. Thus $x^{100}\left[\frac{1}{x}\right] = -x^{100}$ whenever $x \le -1$. What does that tell you about the limit as $x \to -\infty$?

When $-1 \le 1/x < 0$, we have for example $-0,000111$. Isn't the integer part again $0$ ? (Wondering)

 

[Euge]

No, because the integer $0$ is greater than that number. The integer part of a number $z$ is defined as the greatest integer less than or equal to $z$.

 

[mathmari]

No, because the integer $0$ is greater than that number. The integer part of a number $z$ is defined as the greatest integer less than or equal to $z$.

Ah... I though that the integer part of a number of the form "a,bcd..." is "a" so the number before the comma... (Thinking)