148.

6. Granična vrednost tipa: ∞∞ \frac{\infin}{\infin}

TEKST ZADATKA

Odrediti graničnu vrednost:

lim⁡x→+∞(3x22x+1−(2x−1)(3x2+x+2)4x2)\lim_{{x} \to {+\infty}} \bigg(\frac {3x^2} {2x+1} - \frac {(2x-1)(3x^2+x+2)} {4x^2} \bigg)
REŠENJE ZADATKA

Uvrstiti x→∞. x \to \infin . Granična vrednost je neodređenog oblika ∞∞.\dfrac{\infty}{\infty}.

lim⁡x→+∞3x22x+1=3∗∞22∗∞+1=∞∞,lim⁡x→+∞(2x−1)(3x2+x+2)4x2=(2∞−1)(3∞2+∞+2)4∞2=∞∞\lim_{{x} \to {+\infty}} \frac {3x^2} {2x+1} = \frac {3*\infty^2} {2*\infty+1} = \frac {\infty} {\infty} , {\lim_{{x} \to {+\infty}} \frac {(2x-1)(3x^2+x+2)} {4x^2} = \frac {(2\infty-1)(3\infty^2+\infty+2)} {4\infty^2} = \frac {\infty} {\infty}}

Srediti izraz.

lim⁡x→∞(3x22x+1−(2x−1)(3x2+x+2)4x2)=lim⁡x→∞(3x22x+1−2x∗3x2+2x∗x+2x∗2−3x2−x−24x2)=lim⁡x→∞(3x22x+1−6x3+2x2+4x−3x2−x−24x2)=lim⁡x→∞(3x22x+1−6x3−x2+3x−24x2)\lim_{{x} \to {\infty}} \bigg(\frac {3x^2} {2x+1} - \frac {(2x-1)(3x^2+x+2)} {4x^2} \bigg) = \lim_{{x} \to {\infty}} \bigg(\frac {3x^2} {2x+1} - \frac {2x*3x^2+2x*x+2x*2-3x^2-x-2} {4x^2} \bigg)= \lim_{{x} \to {\infty}} \bigg(\frac {3x^2} {2x+1} - \frac {6x^3+2x^2+4x-3x^2-x-2} {4x^2} \bigg) =\lim_{{x} \to {\infty}} \bigg(\frac {3x^2} {2x+1} - \frac {6x^3-x^2+3x-2} {4x^2} \bigg)
lim⁡x→+∞3x2∗4x2−(6x3−x2+3x−2)(2x+1)4x2(2x+1)=lim⁡x→+∞12x4−(6x3∗2x+6x3−x2∗2x−x2+3x∗2x+3x−2∗2x−2)4x2∗2x+4x2)=lim⁡x→+∞12x4−(12x4+6x3−2x3−x2+6x2+3x−4x−2)8x3+4x2\lim_{{x} \to {+\infty}} \frac {3x^2*4x^2-(6x^3-x^2+3x-2)(2x+1)} {4x^2(2x+1)} = \lim_{{x} \to {+\infty}} \frac {12x^4-(6x^3*2x+6x^3-x^2*2x-x^2+3x*2x+3x-2*2x-2)} {4x^2*2x+4x^2)} = \lim_{{x} \to {+\infty}} \frac {12x^4-(12x^4+6x^3-2x^3-x^2+6x^2+3x-4x-2)} {8x^3+4x^2}

Srediti izraz.

lim⁡x→+∞12x4−(12x4+4x3+5x2−x−2)8x3+4x2=lim⁡x→+∞12x4−12x4−4x3−5x2+x+28x3+4x2=lim⁡x→+∞−4x3−5x2+x+28x3+4x2\lim_{{x} \to {+\infty}} \frac {12x^4-(12x^4+4x^3+5x^2-x-2)} {8x^3+4x^2} = \lim_{{x} \to {+\infty}} \frac {12x^4-12x^4-4x^3-5x^2+x+2} {8x^3+4x^2} = \lim_{{x} \to {+\infty}} \frac {-4x^3-5x^2+x+2} {8x^3+4x^2}

Ispred zagrade izvući najveći stepen x.x.

lim⁡x→+∞x3(−4−5x+1x2+2x3)x3(8+4x) \lim_{{x} \to {+\infty}} \frac {x^3 (-4-\frac 5 x+\frac 1 {x^2}+\frac 2 {x^3}) } {x^3(8+\frac 4 x)}
lim⁡x→+∞x3(−4−5x+1x2+2x3)x3(8+4x)=lim⁡x→+∞−4−5x+1x2+2x38+4x \lim_{{x} \to {+\infty}} \frac {\cancel{x^3} (-4-\frac 5 x+\frac 1 {x^2}+\frac 2 {x^3}) } {\cancel{x^3}(8+\frac 4 x)} = \lim_{{x} \to {+\infty}} \frac {-4-\frac 5 x+\frac 1 {x^2}+\frac 2 {x^3} } {8+\frac 4 x}

Uvrstiti x→∞. x \to \infin .

lim⁡x→∓∞−4−5∓∞+1(∓∞)2+2(∓∞)38+4∓∞=lim⁡x→∓∞−4−5∓∞+1∞2+2(∓∞)38+4∓∞=lim⁡x→∓∞−4−0+0+08+0=lim⁡x→∓∞−48=lim⁡x→∓∞−12=−12 \lim_{{x} \to {\mp\infty}} \frac {-4-\frac 5 {\mp\infty}+\frac 1 {(\mp\infty)^2}+\frac 2 {(\mp\infty)^3} } {8+\frac 4 {\mp\infty}} = \lim_{{x} \to {\mp\infty}} \frac {-4-\cancel{\frac 5 {\mp\infty}}+\cancel{\frac 1 {\infty^2}}+\cancel{\frac 2 {(\mp\infty)^3}} } {8+\cancel{\frac 4 {\mp\infty}}} = \lim_{{x} \to {\mp\infty}} \frac {-4-0+0+0 } {8+0} = \lim_{{x} \to {\mp\infty}} \frac {-4} 8 = \lim_{{x} \to {\mp\infty}} \frac {-1} 2= -\frac 1 2