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 April 15th, 2012, 08:35 AM #1 Newbie   Joined: Apr 2012 Posts: 18 Thanks: 0 calculus doubts Hi guys, please help me with the following questions. 1) lim x-->0 x[1/x] where [.] denotes the greatest integer function. 2) if f "(x) exists, then show that lim h-->0 {f(x+h)-2f(x)+f(x-h)}/h^2 = f "(x) 3) lim x-->0 [x]/x (for this one i'm getting 0 as the RHL and -oo as the RHL. please confirm!)
 April 15th, 2012, 08:54 AM #2 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: calculus doubts 1. If x is integer, then x/[x] lim x -> 0 is 0
 April 15th, 2012, 09:18 AM #3 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: calculus doubts 2. f''(x) = d(dy) / dx^2. d(dy) = d(f(x+h)-f(x)) = d(f(x+h)) - d(f(x)) = f(x+2h) - f(x) - f(x+h)-f(x) Now set x+h = t Then f(t+h)-2f(x)-f(t) = d(dy) As h tense to zero, x tense to t. Thus it is proved
April 15th, 2012, 01:20 PM   #4
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Re: calculus doubts

Quote:
 Originally Posted by mathbalarka 1. If x is integer, then x/[x] lim x -> 0 is 0
I think that this is not true. We have $\lim_{x \to 0} x $\frac{1}{x}$$ where $$\frac{1}{x}$= \frac{1}{x} - \{ \frac{1}{x} \}$ (with $\{ \cdot \}$ I indicate the fractional part); and we also have that $0 \le \{ \frac{1}{x} \} < 1 \qquad \forall x \in \mathbb{R} \setminus \{ 0 \}$.

So $\lim_{x \to 0} x $\frac{1}{x}$= \lim_{x \to 0} x \cdot $$\frac{1}{x} - \{ \frac{1}{x} \}$$=\lim_{x \to 0} (1 - x \{ \frac{1}{x} \} )=1$

 April 15th, 2012, 01:24 PM #5 Newbie   Joined: Apr 2012 From: Padua, Italy Posts: 15 Thanks: 0 Re: calculus doubts You can use a quite similar argument to solve the third question...
 April 15th, 2012, 09:48 PM #6 Newbie   Joined: Apr 2012 Posts: 18 Thanks: 0 Re: calculus doubts Thanks delirium. But what argument would you use for the last question? I got this: lim x-->0 x-{x} whole divided by x Now what should I do? If x is very close to zero, then x={x} isn't it? Why can't I do the following: For x-->0+, [x]=0. So limit =0. For x-->0-, [x]=-1 so the limit is -oo ??
 April 16th, 2012, 05:38 AM #7 Newbie   Joined: Apr 2012 From: Padua, Italy Posts: 15 Thanks: 0 Re: calculus doubts Your limit is: $\lim_{x \to 0} \frac{ $x$ }{x}$; but $$x$$ is defined as the biggest integer smaller than $x$; so, if $x \to 0^{+}$, $[x]$ is steadly . But if $x \to 0^{-}$, $\lim_{x \to 0^{-}} f(x)= - \infty$ so I think that this limit doesn't exist, but I'm not sure. EDIT: I found a confirmation in a www.mat.uniroma2.it/~tauraso/Online1/Lezioni/Lezioni1516.pdf+limite+funzione+parte+intera&hl=it &gl=it&pid=bl&srcid=ADGEESjjkZp9Ce8jFTEo_T8Azz_LLv jpnkjFJWXKyf_0R1V2eGzOAaxsH85U8KlhTJqADCSdG-uwRP7qMTU87L70dG5QKwONYqemqdO8D1hJaSnnAUIM4DGAVkbT b-DIzfEv5PaYhNQk&sig=AHIEtbTJZC4WPuSWE5n0fNd6r1REynK zZQ]small paper of University of Rome[/url] (but it is in italian). So I now confirm what I've said.
April 16th, 2012, 09:50 AM   #8
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Re: calculus doubts

Quote:
 Originally Posted by Delirium Your limit is: $\lim_{x \to 0} \frac{ $x$ }{x}$; but $$x$$ is defined as the biggest integer smaller than $x$; so, if $x \to 0^{+}$, $[x]$ is steadly . But if $x \to 0^{-}$, $\lim_{x \to 0^{-}} f(x)= - \infty$ so I think that this limit doesn't exist, but I'm not sure. EDIT: I found a confirmation in a www.mat.uniroma2.it/~tauraso/Online1/Lezioni/Lezioni1516.pdf+limite+funzione+parte+intera&hl=it &gl=it&pid=bl&srcid=ADGEESjjkZp9Ce8jFTEo_T8Azz_LLv jpnkjFJWXKyf_0R1V2eGzOAaxsH85U8KlhTJqADCSdG-uwRP7qMTU87L70dG5QKwONYqemqdO8D1hJaSnnAUIM4DGAVkbT b-DIzfEv5PaYhNQk&sig=AHIEtbTJZC4WPuSWE5n0fNd6r1REynK zZQ]small paper of University of Rome[/url] (but it is in italian). So I now confirm what I've said.
Alright delirium, this is what I was doing too. Thanks a ton
can u help me with question 2 as well?

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