186. Without the l'Hospital's rules.

by Virgil Nicula, Dec 9, 2010, 5:59 PM

I"ll ascertain without l'Hospital's rules the limits of some functions in common hypothesis that these limits exist and are finitely.

$1.\blacktriangleright\ \boxed{\ \lim_{x\to 0}\ \frac {x-\sin x}{x^3}=\frac 16\ }$ .

Proof. Denote $f_1(x)=\frac {x-\sin x}{x^3}\ ,\ x\ne 0$ and $L_1=\lim_{x\to 0}f_1(x)$ (from hypothesis this limit exists and is finitely). Observe that $4\cdot f_1(2x)=f_1(x)+$

$\frac {\sin x}{x}\cdot\frac {1-\cos x}{x^2}$ . For $x\to 0$ obtain that $4L_1=L_1+\frac 12$ $\implies$ $L_1=\frac 16$ . I used only the remarkable limits $\lim_{x\to 0}\frac {\sin x}{x}=1$ and $\lim_{x\to 0}\frac {1-\cos x}{x^2}=\frac 12$ .

Remark. Denote $g(x)=\frac {x-\sin x}{x^2}$ , where $x\ne 0$ . Then prove easily that $\lim_{x\to 0}g(x)=0$ . Indeed, if $0<x<\frac {\pi}{2}$ , then $\sin x<x<\tan x$ $\implies$

$0<x-\sin x<\tan x-\sin x$ $\implies$ $0<g(x)<\frac {\tan x-\sin x}{x^2}=\tan x\cdot\frac {1-\cos x}{x^2}$ $\implies$ $g(0+)=\lim_{x\searrow 0}g(x)=0$ .

But $g(0-)=\lim_{x\nearrow 0}g(x)=$ $\lim_{x\searrow 0}g(-x)=$ $-\lim_{x\searrow 0}g(x)=0$ . In conclusion, $g(0+)=g(0-)=0$ $\implies$ $\lim_{x\to 0}\ \frac {x-\sin x}{x^2}=0$ .

$2.\blacktriangleright\ \boxed{\ \lim_{x\to 0}\ \frac {x-\ln (x+1)}{x^2}=\frac 12\ }$ .

Proof. Denote $f_2(x)=\frac {x-\ln (x+1)}{x^2}\ ,\ -1<x\ne 0$ and $L_2=\lim_{x\to 0}f_2(x)$ (from hypothesis this limit exists and is finitely).

Observe that $1+2\cdot f_2(x)=(x+2)^2\cdot f_2\left(x^2+2x\right)$ . For $x\to 0$ obtain that $1+2L_2=4L_2$ $\implies$ $L_2=\frac 12$ .

$3.\blacktriangleright\ \boxed{\ \lim_{x\to 0}\ \frac {e^x-x-1}{x^2}=\frac 12\ }$ .

Proof. Denote $f_3(x)=\frac {e^x-x-1}{x^2}\ ,\ x\ne 0$ and $L_3=\lim_{x\to 0}f_3(x)$ (from hypothesis this limit exists and is finitely). Observe that $4\cdot f_3(2x)=$

$2\cdot f_3(x)+\left(\frac {e^x-1}{x}\right)^2$ . For $x\to 0$ obtain that $4L_3=2L_3+1$ $\implies$ $L_3=\frac 12$ . I used only the remarkable limit $\lim_{x\to 0}\frac {e^x-1}{x}=1$ .

$4.\blacktriangleright\ \boxed{\ \lim_{x\to 0}\ \frac {e^x-e^{-x}-2x}{x^3}=\frac 13\ }$ .

Proof. Denote $f_4(x)=\frac {e^x-e^{-x}-2x}{x^3}\ ,\ x\ne 0$ and $L_4=\lim_{x\to 0}f_4(x)$ (from hypothesis this limit exists and is finitely). Observe that $27\cdot f_4(3x)=$

$3\cdot f_4(x)+\left(\frac {e^x-e^{-x}}{x}\right)^3$ . For $x\to 0$ obtain that $27L_4=3L_4+8$ $\implies$ $L_3=\frac 13$ . I used only the remarkable limit $\lim_{x\to 0}\frac {e^x-e^{-x}}{x}=2$ .

$5.\blacktriangleright\ \boxed{\ \lim_{x\to 0}\ \frac {e^x-\frac 12\cdot x^2-x-1}{x^3}=\frac 13\ }$ .

Proof. Denote $f_5(x)=\frac {e^x-\frac 12\cdot x^2-x-1}{x^3}\ ,\ x\ne 0$ and $L_5=\lim_{x\to 0}f_5(x)$ . Observe that $27\cdot f_5(3x)=$ $3\cdot f_5(x)+3\cdot f_3(x)\cdot\frac {e^2-1+x}{x}+$

$\left(\frac {e^x-1}{x}\right)^3$ . For $x\to 0$ obtain that $27L_5=3L_5+3\cdot L_3\cdot 2+1$ $\implies$ $L_5=\frac 16$ . I used only the remarkable limit $\lim_{x\to 0}\frac {e^x-1}{x}=1$ .

$6.\blacktriangleright\ \boxed{\ \lim_{x\to 0}\ \frac {e^x+e^{-x}-x^2-2}{x^4}=\frac {1}{12}\ }$ .

Proof. Denote $f_6(x)=\frac {e^x+e^{-x}-x^2-2}{x^4}\ ,\ x\ne 0$ and $L_6=\lim_{x\to 0}f_6(x)$ . Observe that $256\cdot f_6(4x)=$ $4\cdot f_4(x)\cdot \frac {e^x-e^{-x}+2x}{x}+\left(\frac {e^x-e^{-x}}{x}\right)^4$ .

For $x\to 0$ obtain that $256\cdot L_6=16\cdot L_4+16$ $\implies$ $L_6=\frac {1}{12}$ . I used only the previous limit $L_4=\frac 13$ and the remarkable limit $\lim_{x\to 0}\frac {e^x-e^{-x}}{x}=2$ .
This post has been edited 30 times. Last edited by Virgil Nicula, Nov 22, 2015, 6:26 PM

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7. Some interesting "slicing" problems.
6. Interesting problems from JBMO.
5. Theoretical preliminary for inequalities.
4. Remarkable geometrical inequalities (2).
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  • orzzzzzzzzz

    by mathMagicOPS, Jan 9, 2025, 3:40 AM

  • this css is sus

    by ihatemath123, Aug 14, 2024, 1:53 AM

  • 391345 views moment

    by ryanbear, May 9, 2023, 6:10 AM

  • We need virgil nicula to return to aops, this blog is top 10 all time.

    by OlympusHero, Sep 14, 2022, 4:44 AM

  • :omighty: blog

    by tigerzhang, Aug 1, 2021, 12:02 AM

  • Amazing blog.

    by OlympusHero, May 13, 2021, 10:23 PM

  • the visits tho

    by GoogleNebula, Apr 14, 2021, 5:25 AM

  • Bro this blog is ripped

    by samrocksnature, Apr 14, 2021, 5:16 AM

  • Holy- Darn this is good. shame it's inactive now

    by the_mathmagician, Jan 17, 2021, 7:43 PM

  • godly blog. opopop

    by OlympusHero, Dec 30, 2020, 6:08 PM

  • long blog

    by MrMustache, Nov 11, 2020, 4:52 PM

  • 372554 views!

    by mrmath0720, Sep 28, 2020, 1:11 AM

  • wow... i am lost.

    369302 views!

    -piphi

    by piphi, Jun 10, 2020, 11:44 PM

  • That was a lot! But, really good solutions and format! Nice blog!!!! :)

    by CSPAL, May 27, 2020, 4:17 PM

  • impressive :D
    awesome. 358,000 visits?????

    by OlympusHero, May 14, 2020, 8:43 PM

72 shouts
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About Owner
  • Posts: 7054
  • Joined: Jun 22, 2005
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  • Blog created: Apr 20, 2010
  • Total entries: 456
  • Total visits: 404395
  • Total comments: 37
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