20 L’Hôpitalovo pravilo

L’Hôpitalovo pravilo je recept za računanje limit funkcij z nedoločenimi izrazi z uporabo odvodov.

Prva nedoločena oblika, ki jo lahko izračunamo s tem pravilom, je oblika \left[\dfrac{0}{0}\right].

Izrek 20.1 [10]: Če sta f in g taki funkciji, da velja

  1.  f(\alpha) = g(\alpha) = 0
  2.  ter obstaja taka okolica točke \alpha, v kateri sta f in g odvedljivi in za kateri velja g(x) \neq 0 in g'(x) \neq 0 za vsak x \neq \alpha,
  3. in obstaja limita \displaystyle\lim_{x \to \alpha} \frac{f'(x)}{g'(x)},

potem obstaja tudi limita \displaystyle\lim_{x \to \alpha} \frac{f(x)}{g(x)} in velja \displaystyle\lim_{x \to \alpha} \frac{f(x)}{g(x)} =\lim_{x \to \alpha} \frac{f'(x)}{g'(x)}.

Primer 20.1: Izračunajmo naslednje limite:

  1. \displaystyle\lim_{x \to 1} \frac{x^3 - 1}{x^2 - 1} \stackrel{\left[\frac{0}{0}\right]}{\underset{\text{L'H}}{=}} \lim_{x \to 1} \frac{(x^3-1)'}{(x^2-1)'}= \lim_{x \to 1} \frac{3x^2}{2x} = \frac{3}{2}.
  2. \displaystyle\lim_{x \to 0} \frac{e^x-1}{\ln (x+1)} \stackrel{\left[\frac{0}{0}\right]}{\underset{\text{L'H}}{=}} \lim_{x \to 0} \frac{(e^x-1)'}{(\ln (x+1))'}= \lim_{x \to 0} \frac{e^x}{\dfrac{1}{x+1}} = 1.
  3. \displaystyle\lim_{x \to 0} \frac{e^x - 1 - x}{x^2} \stackrel{\left[\frac{0}{0}\right]}{\underset{\text{L'H}}{=}} \lim_{x \to 0} \dfrac{(e^x - 1 - x)'}{(x^2)'}
    = \displaystyle\lim_{x \to 0} \frac{e^x - 1 }{2x} \stackrel{\left[\frac{0}{0}\right]}{\underset{\text{L'H}}{=}} \lim_{x \to 0} \frac{(e^x - 1)' }{(2x)'}= \lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2}.
  4. \displaystyle\lim_{x \to 0} \frac{\sin x}{x}\stackrel{\left[\frac{0}{0}\right]}{\underset{\text{L'H}}{=}} \lim_{x \to 0} \frac{(\sin x)'}{x'}= \lim_{x \to 0} \frac{(\cos x)}{1}=1.
  5. \displaystyle\lim_{x \to 0} \frac{\sin x - x}{x^3} \stackrel{\left[\frac{0}{0}\right]}{\underset{\text{L'H}}{=}}\lim_{x \to 0} \frac{(\sin x - x)'}{(x^3)'}=\lim_{x \to 0} \frac{\cos x - 1}{3x^2}\stackrel{\left[\frac{0}{0}\right]}{\underset{\text{L'H}}{=}}
    =\displaystyle\lim_{x \to 0} \frac{-\sin x }{6x}\stackrel{\left[\frac{0}{0}\right]}{\underset{\text{L'H}}{=}}\lim_{x \to 0} \frac{-\cos x }{6}= -\frac{1}{6}.
  6. \displaystyle\lim_{x \to 0} \frac{2 \sin x - \sin (2x) }{x - \sin x} \stackrel{\left[\frac{0}{0}\right]}{\underset{\text{L'H}}{=}} \lim_{x \to 0} \frac{(2 \sin x - \sin (2x) )'}{(x - \sin x)'}= \lim_{x \to 0} \frac{2 \cos x - 2 \cos (2x) }{1 - \cos x} \stackrel{\left[\frac{0}{0}\right]}{\underset{\text{L'H}}{=}}
    =\displaystyle\lim_{x \to 0}\frac{(2 \cos x - 2 \cos (2x))' }{(1 - \cos x)'} = \lim_{x \to 0} \frac{-2 \sin x + 2 \cdot 2 \cdot \sin (2x)}{ \sin x}
    =\displaystyle \lim_{x \to 0} \frac{-2 \sin x + 2 \cdot 2 \cdot 2\cdot \sin x\cdot \cos x}{ \sin x}= \lim_{x \to 0} (-2+8\cos x) = 6.

Podobno lahko l’Hôpitalovo pravilo uporabimo tudi za izračun limit oblike \left[\dfrac{\infty}{\infty}\right].

Izrek 20.2 [10]: Naj bosta funkciji f in g odvedljivi v okolici točke \alpha in \displaystyle\lim_{x \to \alpha} f(x) {=} \lim_{x \to \alpha} g(x){ =}{\pm \infty}. Če obstaja \displaystyle\lim_{x \to \alpha} \frac{f'(x)}{g'(x)}, potem obstaja tudi limita \displaystyle\lim_{x \to \alpha} \frac{f(x)}{g(x)} in obe limiti sta enaki.

Primer 20.2: Izračunajmo naslednje limite:

  1. \displaystyle\lim_{x \to +\infty} \frac{5x^3+x^2 + 1}{(x+1)^3} \stackrel{\left[\frac{\infty}{\infty}\right]}{\underset{\text{L'H}}{=}} \lim_{x \to +\infty}\frac{(5x^3+x^2 + 1)'}{((x+1)^3)'}{=} \lim_{x \to +\infty} \frac{15x^2+2x}{3\cdot (x+1)^2} \stackrel{\left[\frac{\infty}{\infty}\right]}{\underset{\text{L'H}}{=}}
    =\displaystyle\lim_{x \to +\infty} \frac{(15x^2+2x)'}{(3\cdot (x+1)^2)'} {= } \lim_{x \to +\infty}\frac{30x+2}{6\cdot (x+1)} \stackrel{\left[\frac{\infty}{\infty}\right]}{\underset{\text{L'H}}{=}}\lim_{x \to +\infty}\frac{(30x+2)'}{(6 \cdot(x+1))'}{=}\frac{30}{6} {=} 5.
  2. \displaystyle\lim_{x \to +\infty} \dfrac{x^2}{e^{x}} \stackrel{\left[\frac{\infty}{\infty}\right]}{\underset{\text{L'H}}{=}} \lim_{x \to +\infty} \dfrac{(x^2)'}{(e^{x})'} = \lim_{x \to +\infty} \dfrac{2x}{e^{x}} \stackrel{\left[\frac{\infty}{\infty}\right]}{\underset{\text{L'H}}{=}}
    \displaystyle\lim_{x \to +\infty} \dfrac{(2x)'}{(e^{x})'} = \lim_{x \to +\infty} \dfrac{2}{e^{x}} = \dfrac{2}{+\infty} = 0.
  3. \displaystyle\lim_{x \to +\infty} \dfrac{\ln (\dfrac{x}{x+1})}{\ln (\dfrac{x}{x-1})}\stackrel{\left[\frac{\ln 1}{\ln 1}=\frac{\infty}{\infty}\right]}{\underset{\text{L'H}}{=}}\lim_{x \to +\infty} \dfrac{\left[\ln (\dfrac{x}{x+1})\right]'}{\left[\ln (\dfrac{x}{x-1})\right]'}=
    \displaystyle\lim_{x \to +\infty} \dfrac{\dfrac{1}{\dfrac{x}{x+1}}\cdot \left(\dfrac{x}{x+1}\right)'}{\dfrac{1}{\dfrac{x}{x-1}}\cdot \left(\dfrac{x}{x-1}\right)'}{=}\lim_{x \to +\infty} \dfrac{\dfrac{x+1}{x}\cdot \dfrac{x'\cdot (x+1)-x\cdot (x+1)'}{(x+1)^2}}{\dfrac{x-1}{x}\cdot \dfrac{x'\cdot (x-1)-x\cdot (x-1)'}{(x-1)^2}}
    {=}\displaystyle\lim_{x \to +\infty} \dfrac{\dfrac{1\cdot (x+1-x)}{x\cdot (x+1)}}{\dfrac{1\cdot (x-1-x)}{x\cdot (x-1)}}{=}\lim_{x \to +\infty} \dfrac{\dfrac{1}{x \cdot (x+1)}}{\dfrac{-1}{x\cdot (x-1)}} {=} \lim_{x \to +\infty} \dfrac{-x\cdot (x-1)}{x\cdot (x+1)}{=}-1.
  4. \displaystyle\lim_{x \to 0^+} \dfrac{\ln (x^2+2x)}{\ln x} \stackrel{\left[\frac{\infty}{\infty}\right]}{\underset{\text{L'H}}{=}}\lim_{x \to 0^+} \dfrac{(\ln (x^2+2x))'}{(\ln x)'}= \lim_{x \to 0^+} \dfrac{\dfrac{2x+2}{x^2+2x}}{\dfrac{1}{x}}
    =\displaystyle \lim_{x \to 0^+} x\cdot \dfrac{2x+2}{x\cdot (x+2)}=\lim_{x \to 0^+} \dfrac{2x+2}{x+2}=\dfrac{2}{2}=1.
  5. \displaystyle\lim_{x \to 0^+} \dfrac{\ln (e^x-1)}{\ln x} \stackrel{\left[\frac{\infty}{\infty}\right]}{\underset{\text{L'H}}{=}}\lim_{x \to 0^+} \dfrac{(\ln (e^x-1))'}{(\ln x)'} =\lim_{x \to 0^+} \dfrac{\dfrac{e^x}{e^x - 1}}{\dfrac{1}{x}}
    =\displaystyle\lim_{x \to 0^+} \dfrac{x \cdot e^x}{e^x - 1} \stackrel{\left[\frac{0}{0}\right]}{\underset{\text{L'H}}{=}} =\lim_{x \to 0^+} \dfrac{(x \cdot e^x)'}{(e^x - 1)'} =
    =\displaystyle\lim_{x \to 0^+} \dfrac{x'\cdot e^x + x \cdot (e^x)'}{e^x } = \lim_{x \to 0^+} \dfrac{e^x + x \cdot e^x}{e^x } = \dfrac{1}{1}=1.

Nedoločeno obliko [\infty \cdot 0] ali [\infty-\infty] najprej preoblikujemo v ulomek \left[\dfrac{0}{0}\right] ali \left[\dfrac{\infty}{\infty}\right] in potem uporabimo l’Hôpitalovo pravilo.

Primer 20.3: Izračunajmo naslednje limite:

  1. \displaystyle\lim_{x \to 0} \left(\dfrac{1}{\sin x} - \dfrac{1}{x}\right) \stackrel{\left[\infty-\infty\right]}{=} \lim_{x \to 0} \dfrac{x-\sin x}{x\cdot \sin x}\stackrel{\left[\frac{0}{0}\right]}{\underset{\text{L'H}}{=}} \lim_{x \to 0}\dfrac{(x-\sin x)'}{(x\cdot \sin x)'}
    = \displaystyle\lim_{x \to 0} \dfrac{1 - \cos x}{\sin x + x \cdot \cos x} \stackrel{\left[\frac{0}{0}\right]}{\underset{\text{L'H}}{=}} \lim_{x \to 0} \dfrac{(1 - \cos x)'}{(\sin x + x \cdot \cos x)'}
    = \displaystyle\lim_{x \to 0} \dfrac{\sin x}{\cos x+\cos x + x \cdot (-\sin x)} = \dfrac{0}{2} = 0.
  2. \displaystyle\lim_{x \to 1^+} \left(\dfrac{1}{x-1} - \frac{1}{\ln x}\right) \stackrel{\left[\infty-\infty\right]}{=} \lim_{x \to 1^+} \dfrac{\ln x - (x-1)}{(x-1) \cdot \ln x} \stackrel{\left[\frac{0}{0}\right]}{\underset{\text{L'H}}{=}} \lim_{x \to 1^+} \dfrac{(\ln x - x+1)'}{((x-1) \cdot \ln x)'}=\displaystyle\lim_{x \to 1^+} \dfrac{\dfrac{1}{x} -1}{\ln x +(x-1)\cdot\dfrac{1}{x}} \stackrel{\left[\frac{0}{0}\right]}{\underset{\text{L'H}}{=}} \lim_{x \to 1^+} \dfrac{\left(\dfrac{1}{x} -1\right)'}{\left(\ln x +1-\dfrac{1}{x}\right)'}
    =\displaystyle\lim_{x \to 1^+} \dfrac{-x^{-2}}{\dfrac{1}{x}+x^{-2}} = -\dfrac{1}{2}.
  3. \displaystyle\lim_{x \to 0^+} (\sqrt{x} \cdot \ln x) \stackrel{\left[0\cdot(-\infty)\right]}{=}\lim_{x \to 0^+} \dfrac{\ln x}{\dfrac{1}{\sqrt{x}}}\stackrel{\left[\frac{\infty}{\infty}\right]}{\underset{\text{L'H}}{=}}\lim_{x \to 0^+} \dfrac{(\ln x)'}{(\dfrac{1}{\sqrt{x}})'}
    =\displaystyle\lim_{x \to 0^+} \dfrac{\dfrac{1}{x}}{-\dfrac{1}{x}\cdot \dfrac{1}{2\sqrt x}}= \lim_{x \to 0^+} - \dfrac{2x\sqrt{x}}{x} = \lim_{x \to 0^+} (- 2\sqrt{x})=0.
  4. \displaystyle\lim_{x \to 0^+} (x^2 \cdot \ln x) \stackrel{\left[0\cdot(-\infty)\right]}{=} \lim_{x \to 0^+} \dfrac{\ln x}{\dfrac{1}{x^2}}\stackrel{\left[\frac{\infty}{\infty}\right]}{\underset{\text{L'H}}{=}} \lim_{x \to 0^+} \dfrac{(\ln x)'}{(\dfrac{1}{x^2})'}
    =\displaystyle\lim_{x \to 0^+} \dfrac{\dfrac{1}{x}}{\dfrac{-2}{x^3}} = \lim_{x \to 0^+} -\frac{x^2}{2} = 0.
  5. \displaystyle\lim_{x \to 0^+} (x^3 \cdot e^{\frac{1}{x}}) \stackrel{\frac{1}{x}=y}{=} \lim_{y \to +\infty} \dfrac{1}{y^3}\cdot e^y=\lim_{y \to +\infty} \dfrac{e^y}{y^3}\stackrel{\left[\frac{\infty}{\infty}\right]}{\underset{\text{L'H}}{=}}\lim_{y \to +\infty} \dfrac{e^y}{3y^2}\stackrel{\left[\frac{\infty}{\infty}\right]}{\underset{\text{L'H}}{=}}
    =\displaystyle\lim_{y \to +\infty} \dfrac{e^y}{6y}\stackrel{\left[\frac{\infty}{\infty}\right]}{\underset{\text{L'H}}{=}} \lim_{y \to +\infty} \dfrac{e^y}{6}=+\infty.

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