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13 Naloge s celotnim postopkom reševanja

Naloga 1:  Določite vrednosti a in b tako, da bo funkcija f: \mathbb{R} \longrightarrow \mathbb{R}, podana s predpisom

    \[f(x) = \begin{cases} \dfrac{x^2-4}{x-2}, &\textup{če je }x < 2, \\ ax^2-bx+3, &\textup{če je }2 \leq x < 3,\\ 2x-a+b,&\textup{če je }x\geq 3, \end{cases}\]

zvezna na \mathbb{R}.

Rešitev: Funkcija f je zvezna na \mathbb{R}, če je zvezna v vsaki točki, torej tudi v točkah x=2 in x=3. Leva limita v točki 2 je

    \[\lim_{x \to 2^-} f(x)=\lim_{x \to 2^-} \frac{x^2-4}{x-2}=\lim_{x \to 2^-} \frac{(x-2)\cdot (x+2)}{x-2}=\lim_{x \to 2^-} (x+2)=4,\]

desna pa je

    \[\lim_{x \to 2^+} f(x)= \lim_{x \to 2^+} (ax^2-bx+3)= 4a-2b+3= f(2).\]

Funkcija f je v točki 2 zvezna, če velja

    \[\lim_{x \to 2^-} f(x)= \lim_{x \to 2^+} f(x)= f(2),\]

ekvivalentno z

    \[4=4a-2b+3 \Longleftrightarrow 4a-2b=1.\]

Podobno razpravljamo o zveznosti funkcije f v točki 3. Leva limita v točki 3 je

    \[\lim_{x \to 3^-} f(x)=\lim_{x \to 3^-}(ax^2-bx+3)=9a-3b+3,\]

desna pa je

    \[\lim_{x \to 3^+} f(x)= \lim_{x \to 3^+} (2x-a+b)= 6-a+b= f(3).\]

Funkcija f je v točki 3 zvezna, če velja

    \[\lim_{x \to 3^-} f(x)= \lim_{x \to 3^+} f(x)= f(3)\]

oziroma

    \[9a-3b+3=6-a+b \Longleftrightarrow 10a-4b=3.\]

Zdaj moramo rešiti sistem

    \[\begin{cases} 4a-2b=1\\ 10a-4b=3 \end{cases},\]

ki ima rešitev a=b=\dfrac{1}{2}.

Naloga 2: Naj bo funkcija f definirana s predpisom

    \[f(x)=\begin{cases} \sqrt{x-4},&\textup{\v ce je }x > 4, \\ 8 - 2x,&\textup{\v ce je }x \leq 4.\end{cases}\]

Ali obstaja limita \displaystyle\lim_{x \to 4} f(x)?

Rešitev: Izračunamo levo in desno limito funkcije f v točki x=4:
\displaystyle\lim_{x \to 4^-} f(x)= &\lim_{x \to 4^-} (8-2x)= 8-2\cdot 4=0

\displaystyle\lim_{x \to 4^+} f(x)= &\Lim_{x \to 4^+} \sqrt{x-4}=0.

Torej, funkcija f ima limito v točki 4, in sicer \displaystyle\lim_{x \to 4} f(x)=0.

Naloga 3: Ali obstaja limita \displaystyle\lim_{x \to 0} \dfrac{5}{3-e^{\frac{1}{x}}}?

Rešitev: Funkcija f(x)=\dfrac{1}{x} ni definirana v točki x=0 (v tej točki ima pol), torej je \displaystyle\lim_{x \to 0^+} \frac{1}{x}=\frac{1}{0^+}=+\infty ter \displaystyle\lim_{x \to 0^-} \frac{1}{x}=\frac{1}{0^-}=-\infty. Zato dobimo \displaystyle\lim_{x \to 0^+} e^{\frac{1}{x}}=e^{+\infty}=+\infty ter \displaystyle\lim_{x \to 0^-} e^{\frac{1}{x}}=e^{-\infty}=\frac{1}{e^{+\infty}}=\frac{1}{+\infty}=0^+. To pomeni, da ne obstaja limita \displaystyle\lim_{x \to 0} e^{\frac{1}{x}} in posledično niti ne dana limita.

Naloga 4: Izračunajte limito \displaystyle\lim_{x\to-3}\frac{x^2-9}{2x^2+7x+3}.

Rešitev:
\displaystyle\lim_{x\to-3}\frac{x^2-9}{2x^2+7x+3}&=\lim_{x \to -3} \frac{(x+3)\cdot (x-3)}{2\cdot (x+3)\cdot (x+ \frac{1}{2})}=\displaystyle\frac{1}{2} \lim_{x\to-3}\frac{x-3}{x+\dfrac{1}{2}}

=\displaystyle\frac{1}{2}\cdot\frac{-6}{-\dfrac{5}{2}}=\frac{6}{5}.

Naloga 5: Izračunajte limito \displaystyle\lim_{x \to -1} \frac{x^2 + 2x +1}{x^4 - 1}.

Rešitev:

\displaystyle\lim_{x \to -1} \frac{x^2 + 2x +1}{x^4 - 1} &= \lim_{x \to -1} \frac{(x+1)^2}{(x^2 - 1) \cdot (x^2 + 1)}

=\displaystyle\lim_{x \to -1} \frac{(x+1)^2}{(x - 1) \cdot (x+1) \cdot (x^2 + 1)}

=\displaystyle\lim_{x \to -1} \frac{x+1}{(x-1) \cdot (x^2 + 1)}

= \dfrac{0}{-4} = 0.

Naloga 6: Izračunajte limito \displaystyle\lim_{x \to -\infty} \frac{\sqrt{9x^6 - x} }{x^3 + 1}.

Rešitev:

\displaystyle\lim_{x \to -\infty} \frac{ \sqrt{9x^6 - x} }{x^3 + 1} = \lim_{x \to -\infty} \frac{ \sqrt{x^6 \cdot \left(9-\frac{x}{x^6}\right)} }{x^3 \cdot \left(1 + \frac{1}{x^3}\right)}

=\displaystyle\lim_{x \to -\infty} \frac{ x^3 \cdot \sqrt{9 - \frac{1}{x^5}} }{x^3\cdot\left(1+\frac{1}{x^3}\right)}

=\displaystyle\lim_{x \to -\infty} \frac{\sqrt{9 - \boxed{\frac{1}{x^5}}_{\searrow 0}}}{\left(1+\boxed{\frac{1}{x^3}}_{\searrow 0}\right)}

=\dfrac{\sqrt{9}}{1}=3.

Naloga 7: Izračunajte limito \displaystyle\lim_{x \to +\infty} \left(1+\frac{4}{x}\right)^{x-3}.

Rešitev: Ker je dana limita v nedoločeni obliki [1^{\infty}], uporabimo formulo (1):

\displaystyle\lim_{x \to +\infty} \left(1+\frac{4}{x}\right)^{x-3}{=}\lim_{x \to +\infty} \left[\left(1+\frac{1}{\frac{x}{4}}\right)^{\frac{x}{4}}\right]^{\frac{4}{x}\cdot (x-3)}{=}e^{\displaystyle\lim_{x \to +\infty} \frac{4(x-3)}{x}}{=}e^4.

Naloga 8: Izračunajte limito \displaystyle\lim_{x \to +\infty} \left(\frac{x+6}{x+1}\right)^{x+3}.

Rešitev: Ker je dana limita v nedoločeni obliki [1^{\infty}], uporabimo formulo (1):

\displaystyle\lim_{x \to +\infty} \left(\frac{x+6}{x+1}\right)^{x+3} &=\lim_{x \to +\infty} \left(1+\frac{x+6}{x+1}-1\right)^{x+3}

=\displaystyle \lim_{x \to +\infty} \left(1+\frac{x+6-(x+1)}{x+1}\right)^{x+3}

=\displaystyle \lim_{x \to +\infty} \left(1+ \frac{5}{x+1}\right)^{x+3}

=\displaystyle\lim_{x \to +\infty} \left[\left(1+\frac{1}{\frac{x+1}{5}}\right)^{\frac{x+1}{5}}\right]^{\frac{5}{x+1} \cdot (x+3)}

= e^{\displaystyle\lim_{x \to +\infty} \frac{5(x+3)}{x+1}}

= e^5.

Naloga 9: Izračunajte limito \displaystyle\lim_{x \to +\infty} \left(\frac{x+1}{x-1}\right)^x.

Rešitev: Ker je dana limita v nedoločeni obliki [1^{\infty}], uporabimo formulo (1):

\displaystyle\lim_{x \to +\infty} \left(\frac{x+1}{x-1}\right)^x&=\lim_{x \to +\infty} \left(1+\frac{x+1}{x-1}-1\right)^x

=\displaystyle\lim_{x \to +\infty} \left(1+\frac{x+1-x+1}{x-1}\right)^x

=\displaystyle\lim_{x \to +\infty} \left(1+\frac{2}{x-1}\right)^x

=\displaystyle \lim_{x \to +\infty} \left(1+\frac{1}{\frac{x-1}{2}}\right)^x = \lim_{x \to +\infty} \left[\left(1+\frac{1}{\frac{x-1}{2}}\right)^{\frac{x-1}{2}} \right]^{\frac{2x}{x-1}}

= e^{\displaystyle\lim_{x \to +\infty} \frac{2x}{x-1}}

= e^2.

Naloga 10: Izračunajte limito \displaystyle\lim_{x \to 0} \sqrt[x]{1-5x}.

Rešitev: Najprej dano limito zapišemo v obliki

    \[\lim_{x \to 0} \sqrt[x]{1-5x} =\lim_{x \to 0} (1-5x)^{\frac{1}{x}},\]

torej v nedoločeni obliki [1^{\infty}], zato lahko uporabimo formulo (1):

    \[\lim_{x \to 0} (1-5x)^{\frac{1}{x}}=\lim_{x \to 0} \left[\left(1 + \frac{1}{-\frac{1}{5x}} \right)^{-\frac{1}{5x}} \right]^{-\frac{5x}{x}} = e^{-5}.\]

Naloga 11: Izračunajte limito \displaystyle\lim_{x \to -\infty} (\sqrt{4x^2-x+3} +2x).

Rešitev: Dana limita je v nedoločeni obliki [\infty-\infty], torej najprej uporabimo metodo razširjanja.

\displaystyle \lim_{x \to -\infty} (\sqrt{4x^2-x+3} +2x)&=\lim_{x \to -\infty} (\sqrt{4x^2-x+3} +2x)\cdot \frac{\sqrt{4x^2-x+3} -2x}{\sqrt{4x^2-x+3} -2x}

=\displaystyle\lim_{x \to -\infty} \frac{4x^2-x+3-4x^2}{\sqrt{4x^2-x+3} -2x}

=\displaystyle\lim_{x \to -\infty} \frac{-x\cdot (1-\dfrac{3}{x})}{\sqrt{x^2\cdot \left(4-\dfrac{x}{x^2}+\dfrac{3}{x^2}\right)}-2x}

=\displaystyle\lim_{x \to -\infty} \frac{-x\cdot \left(1-\boxed{\frac{3}{x}}^{\nearrow 0}\right)}{-x\cdot \left(\sqrt{4-\boxed{\frac{1}{x}}_{\searrow 0}+\boxed{\frac{3}{x^2}}_{\searrow 0}}+2\right)}

=\dfrac{1}{4},

kjer smo za x \to -\infty upoštevali, da je \sqrt{x^2}=\lvert x\rvert =-x.

Naloga 12: Izračunajte limito \displaystyle\lim_{x \to +\infty} \sqrt{x^2 + 4x +1}-x.

Rešitev: Dana limita je v nedoločeni obliki [\infty-\infty], torej najprej uporabimo metodo razširjanja.

\displaystyle\lim_{x \to +\infty} (\sqrt{x^2 + 4x +1}-x)&=\lim_{x \to +\infty} ( \sqrt{x^2 + 4x +1}-x)\cdot \frac{ \sqrt{x^2 + 4x +1}+x}{ \sqrt{x^2 + 4x +1}+x}

=\displaystyle\lim_{x \to +\infty} \frac{ x^2 + 4x +1-x^2}{ \sqrt{x^2 + 4x +1}+x}

=\displaystyle\lim_{x \to +\infty} \frac{x\cdot (4+\frac{1}{x})}{ \sqrt{x^2 + 4x +1}+x}

=\displaystyle\lim_{x \to +\infty} \frac{x\cdot\left(4+\boxed{\dfrac{1}{x}}^{\nearrow 0}\right)}{\lvert x\lvert \cdot\left(\sqrt{1+\boxed{\frac{4}{x}}_{\searrow 0}+\boxed{\frac{1}{x^2}}_{\searrow 0}}+1\right)}

=\dfrac{4}{2}=2,

kjer smo za x \to +\infty upoštevali, da je \sqrt{x^2}=\lvert x\rvert =x.

Naloga 13: Izračunajte limito \displaystyle\lim_{x \to -\infty} (\sqrt{x^2-x+2} +x).

Rešitev: Dana limita je v nedoločeni obliki [\infty-\infty], torej najprej uporabimo metodo razširjanja.

\displaystyle\lim_{x \to -\infty} (\sqrt{x^2-x+2} +x)=\lim_{x \to -\infty} (\sqrt{x^2-x+2} +x)\cdot \frac{\sqrt{x^2-x+2} -x}{\sqrt{x^2-x+2} -x}

=\displaystyle\lim_{x \to -\infty} \frac{x^2-x+2-x^2}{\sqrt{x^2-x+2} -x}

=\displaystyle\lim_{x \to -\infty} \frac{-x+2}{\sqrt{x^2\cdot \left(1-\dfrac{x}{x^2}+\dfrac{2}{x^2}\right)}-x}

=\displaystyle\lim_{x \to -\infty} \frac{-x+2}{-x\cdot\sqrt{1-\dfrac{1}{x}+\dfrac{2}{x^2}}-x}

=\displaystyle\lim_{x \to -\infty} \frac{x\cdot\left(-1+\boxed{\frac{2}{x}}^{\nearrow 0}\right)}{x\cdot\left(-\sqrt{1-\boxed{\frac{1}{x}}_{\searrow 0}+\boxed{\frac{2}{x^2}}_{\searrow 0}}-1\right)}

=\dfrac{-1}{-2}=\dfrac{1}{2},

kjer smo za x \to -\infty upoštevali, da je \sqrt{x^2}=\lvert x\rvert =-x.

Naloga 14: Izračunajte limito \displaystyle\lim_{x \to 0}\frac{\sin (4x)}{\sqrt{x+1} -1}.

Rešitev: Najprej uporabimo metodo razširjanja:

\displaystyle\lim_{x \to 0}\frac{\sin (4x)}{\sqrt{x+1} -1} &= \lim_{x \to 0} \frac{\sin (4x)}{\sqrt{x+1} -1} \cdot \frac{\sqrt{x+1} + 1}{\sqrt{x+1} + 1}

= \displaystyle\lim_{x \to 0} \frac{\sin (4x)\cdot \left(\sqrt{x+1} +1\right)} {x+1 - 1}

=\displaystyle \lim_{x \to 0} \frac{\sin (4x)}{x}\cdot \frac{\sqrt{x+1} +1}{1}

=\displaystyle \lim_{x \to 0} 4 \cdot \boxed{\frac{\sin (4x)}{4x}}_{\searrow 1}\cdot\left(\sqrt{x+1} +1\right)

= 4 \cdot 2 = 8.

Naloga 15: Izračunajte limito \displaystyle\lim_{x \to 0}\frac{x + x\cdot \cos x}{\sin x\cdot \cos x}.

Rešitev:

\displaystyle\lim_{x \to 0}\frac{x + x\cdot \cos x}{\sin x\cdot \cos x} = \lim_{x \to 0}\frac{x (1+\cos x)}{\sin x\cdot \cos x}

=\displaystyle\lim_{x \to 0} \frac{x}{\sin x} \cdot \frac{1+\cos x}{\cos x}

=\displaystyle\lim_{x \to 0} \frac{1}{\boxed{\frac{\sin x}{x}}_{\searrow 1}}\cdot\frac{1+\cos x}{\cos x}

=1\cdot2=2.

Naloga 16: Poiščite konstanti a in b tako, da velja

    \[\lim_{x \to 0} \frac{\sqrt{a+bx}-1}{x} = 2.\]

Rešitev: Ko gre x proti 0, je imenovalec 0, torej mora biti tudi števec 0, da dobimo nedoločeno obliko [\frac{0}{0}] (sicer bo limita enaka \infty). To pomeni, da je \sqrt{a+bx} -1 =0, ko gre x proti 0. Dobimo \sqrt{a+0} = 1 in zato je a=1.

Vstavimo vrednost a = 1 v dano funkcijo in dobimo

\displaystyle\lim_{x \to 0} \frac{\sqrt{a+bx}-1}{x} = \lim_{x \to 0} \frac{\sqrt{1+bx}-1}{x} = \lim_{x \to 0} \frac{\sqrt{1+bx}-1}{x} \cdot \frac{\sqrt{1+bx}+1}{\sqrt{1+bx}+1}

{=}\displaystyle\lim_{x \to 0} \frac{1+bx -1}{x \cdot ( \sqrt{1+bx}+1)} {=} \lim_{x \to 0} \frac{bx }{x \cdot( \sqrt{1+bx}+1)} {=} \lim_{x \to 0} \frac{b}{\sqrt{1+bx}+1} {=} \frac{b}{2}.

Ker je po navodilu vrednost limite enaka 2, mora biti \dfrac{b}{2} = 2 oziroma b = 4.

Naloga 17: Izračunajte limito \displaystyle\lim_{x \to 3} \frac{\sqrt{x+6}-x}{x^3-3x^2}.

Rešitev: Dana limita je v nedoločeni obliki [\frac{0}{0}], torej najprej uporabimo metodo razširjanja.

\displaystyle\lim_{x \to 3} \frac{\sqrt{x+6}-x}{x^3-3x^2}=\lim_{x \to 3} \frac{\sqrt{x+6}-x}{x^3-3x^2}\cdot \frac{ \sqrt{x+6}+x}{ \sqrt{x+6}+x}=\lim_{x \to 3} \frac{x+6-x^2}{(x^3-3x^2)\cdot( \sqrt{x+6}+x)}

=\displaystyle\lim_{x \to 3} \frac{-(x-3)\cdot (x+2)}{x^2\cdot (x-3)\cdot (\sqrt{x+6}+x)}= \lim_{x \to 3} \frac{-(x+2)}{x^2\cdot (\sqrt{x+6}+x)} = -\frac{5}{54}.

Naloga 18: Izračunajte limito \displaystyle\lim_{x \to -\infty} \sqrt{x^2+3x} - \sqrt{x^2-2x}.

Rešitev: Dana limita je v nedoločeni obliki [\infty - \infty], torej najprej uporabimo metodo razširjanja.

\displaystyle\lim_{x \to -\infty} \sqrt{x^2+3x} - \sqrt{x^2-2x}&=\lim_{x \to -\infty} (\sqrt{x^2+3x} - \sqrt{x^2-2x})\cdot \frac{ \sqrt{x^2+3x} + \sqrt{x^2-2x}}{\sqrt{x^2+3x} +\sqrt{x^2-2x}}

=\displaystyle\lim_{x \to -\infty} \frac{ x^2 + 3x -x^2 + 2x}{ \sqrt{x^2+3x} + \sqrt{x^2-2x}}

=\displaystyle\lim_{x \to -\infty} \frac{5x}{ \sqrt{x^2\cdot \left(1+ \dfrac{3}{x}\right)}+ \sqrt{x^2\cdot(1-\dfrac{2}{x})}}

=\displaystyle\lim_{x \to -\infty} \frac{5x}{\lvert x\lvert \cdot\left(\sqrt{1+\boxed{\frac{3}{x}}_{\searrow 0}}\right) +\lvert x\lvert \cdot\left(\sqrt{1-\boxed{\frac{2}{x}}_{\searrow 0}}\right)}

=\displaystyle\lim_{x \to -\infty} \frac{5x}{-2x}

=-\dfrac{5}{2},

kjer smo za x \to -\infty upoštevali, da je \sqrt{x^2}=\lvert x\rvert =-x.

Naloga 19: Izračunajte limito \displaystyle\lim_{x \to 9} \frac{\sin (\sqrt{x}-3)}{x-9}.

Rešitev:

    \[\lim_{x \to 9} \frac{\sin (\sqrt{x}-3)}{x-9} {=} \lim_{x \to 9} \frac{\sin (\sqrt{x}-3)}{(\sqrt{x}-3)\cdot (\sqrt{x}+3)}{=}\lim_{x \to 9} \frac{ \sin (\sqrt{x}-3)}{\sqrt{x}-3} \cdot \frac{1}{\sqrt{x}+3} {=} 1 \cdot \frac{1}{6} {=} \frac{1}{6}.\]

Naloga 20: Izračunajte limito \displaystyle\lim_{x \to -\infty} \frac{\sqrt[3]{x}-\sqrt[5]{x}}{\sqrt[3]{x}+\sqrt[5]{x}}.

Rešitev:

    \[\lim_{x \to -\infty} \frac{\sqrt[3]{x}-\sqrt[5]{x}} {\sqrt[3]{x}+\sqrt[5]{x}}= \lim_{x \to -\infty} \frac{x^{\frac{1}{3}}-x^{\frac{1}{5}}}{x^{\frac{1}{3}}+x^{\frac{1}{5}}}=\lim_{x \to -\infty} \frac{x^{\frac{1}{3}}\cdot\left(1-\boxed{x^{-\frac{2}{15}}}_{\searrow 0}\right)}{x^{\frac{1}{3}}\cdot\left(1+\boxed{x^{-\frac{2}{15}}}_{\searrow 0}\right)} = 1.\]

Naloga 21: Izračunajte limito \displaystyle\lim_{x \to +\infty} \frac{2\sqrt{x}+x^{-1}}{3x-7}.

Rešitev:

    \[\lim_{x \to +\infty} \frac{2\sqrt{x}+x^{-1}}{3x-7}{=} \lim_{x \to +\infty} \frac{\sqrt{x}\cdot\left(2+\boxed{\frac{1}{x\sqrt{x}}}_{\searrow 0}\right)}{x\cdot \left(3-\boxed{\frac{7}{x}}_{\searrow 0}\right)}{=}\lim_{x \to +\infty} \frac{2\sqrt{x}}{3x} {=}\lim_{x \to +\infty} \frac{2}{3\sqrt{x}} {=} 0.\]

 

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