Limes ∞/∞: razlika dva racionalna izraza

TEKST ZADATKA

Odrediti graničnu vrednost:

\lim_{{x} \to {+\infty}} \bigg(\frac {3x^2} {2x+1} - \frac {(2x-1)(3x^2+x+2)} {4x^2} \bigg)

REŠENJE ZADATKA

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

\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} \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} \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} \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 $x.$

\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} \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 $x \to \infin .$

\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

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