189.

Trigonometrijski limes

TEKST ZADATKA

Odrediti graničnu vrednost:

lim⁡x→αsin⁡2x−sin⁡2αx2−α2\lim_{{x} \to {\alpha}} \frac {\sin^2{x}-\sin^2{\alpha}} {x^2-{\alpha}^2}
REŠENJE ZADATKA

Primeniti formulu za sinus poluugla: ∣sin⁡α2∣=1−cos⁡α2|\sin{\frac {\alpha} 2}|=\sqrt{\frac {1-\cos{\alpha}} 2}

lim⁡x→α1−cos⁡2x2−1−cos⁡2α2x2−α2\lim_{{x} \to {\alpha}} \frac {\frac {1-\cos{2x}} 2-\frac {1-\cos{2\alpha}} 2} {x^2-{\alpha}^2}
DODATNO OBJAŠNJENJE

Osloboditi se dvojnog razlomka:

lim⁡x→α1−cos⁡2x−1+cos⁡2α2x2−α2=lim⁡x→αcos⁡2α−cos⁡2x2x2−α2=lim⁡x→αcos⁡2α−cos⁡2x2(x2−α2)\lim_{{x} \to {\alpha}} \frac {\frac {1-\cos{2x}-1+\cos{2\alpha}} 2} {x^2-{\alpha}^2}= \lim_{{x} \to {\alpha}} \frac {\frac {\cos{2\alpha}-\cos{2x}} 2} {x^2-{\alpha}^2}= \lim_{{x} \to {\alpha}} \frac {{\cos{2\alpha}-\cos{2x}} } {2(x^2-{\alpha}^2)}

Primeniti formulu za razliku kvadrata:

lim⁡x→αcos⁡2α−cos⁡2x2(x−α)(x+α) \lim_{{x} \to {\alpha}} \frac {{\cos{2\alpha}-\cos{2x}} } {2(x-\alpha)(x+\alpha)}

Primeniti formulu za transformaciju razlike kosinusa: cos⁡α−cos⁡β=−2sin⁡α+β2sin⁡α−β2\cos{\alpha}-\cos{\beta}=-2\sin{\frac {\alpha+\beta} 2}\sin{\frac {\alpha-\beta} 2}

lim⁡x→α−2sin⁡2α+2x2sin⁡2α−2x22(x−α)(x+α) \lim_{{x} \to {\alpha}} \frac {-2\sin{\frac {2\alpha+2x} 2}\sin{\frac {2\alpha-2x} 2} } {2(x-\alpha)(x+\alpha)}

Izvući zajednički činilac ispred zagrade u sinusu:

lim⁡x→α−2sin⁡2(α+x)2sin⁡2(α−x)22(x−α)(x+α)\lim_{{x} \to {\alpha}} \frac {-2\sin{\frac {2(\alpha+x)} 2}\sin{\frac {2(\alpha-x)} 2} } {2(x-\alpha)(x+\alpha)}

Uvesti minus ispred izraza u brojiocu u sinus:

lim⁡x→α2sin⁡2(α+x)2sin⁡2(x−α)22(x−α)(x+α)\lim_{{x} \to {\alpha}} \frac {2\sin{\frac {2(\alpha+x)} 2}\sin{\frac {2(x-\alpha)} 2} } {2(x-\alpha)(x+\alpha)}

Skratiti zajednički činilac:

lim⁡x→α2sin⁡2(α+x)2sin⁡2(x−α)22(x−α)(x+α)=lim⁡x→αsin⁡(α+x)sin⁡(x−α)(x−α)(x+α)\lim_{{x} \to {\alpha}} \frac {\cancel{2}\sin{\frac {\cancel{2}(\alpha+x)} {\cancel{2}}}\sin{\frac {\cancel{2}(x-\alpha)} {\cancel{2}}} } {\cancel{2}(x-\alpha)(x+\alpha)}=\lim_{{x} \to {\alpha}} \frac {\sin{{(\alpha+x)} }\sin{(x-\alpha)} } {(x-\alpha)(x+\alpha)}

Raščlaniti izraz:

lim⁡x→αsin⁡(α+x)α+x∗lim⁡x→αsin⁡(x−α)x−α\lim_{{x} \to {\alpha}} \frac {\sin{(\alpha+x)} } {\alpha+x}*\lim_{{x} \to {\alpha}} \frac {\sin{(x-\alpha)}} {x-\alpha}
DODATNO OBJAŠNJENJE

Zameniti x=αx=\alpha i konstatovati neodređenost 00.\frac{0}{0}.

sin⁡(α+α)α+α∗sin⁡(α−α)α−α=00\frac {\sin{(\alpha+\alpha)} } {\alpha+\alpha}* \frac {\sin{(\alpha-\alpha)}} {\alpha-\alpha}=\frac 0 0

Uvesti smenu:

x−α=ux-\alpha=u
DODATNO OBJAŠNJENJE

Uvrstiti smenu u izraz:

lim⁡x→αsin⁡(α+x)α+x∗lim⁡u→0sin⁡uu\lim_{{x} \to {\alpha}} \frac {\sin{(\alpha+x)} } {\alpha+x}*\lim_{{u} \to {0}} \frac {\sin{u}}u

Ponovo pokušati zameniti x=α.x=\alpha.

lim⁡x→αsin⁡(α+α)α+α∗lim⁡u→0sin⁡uu=lim⁡x→αsin⁡2α2α∗lim⁡u→0sin⁡uu\lim_{{x} \to {\alpha}} \frac {\sin{(\alpha+\alpha)} } {\alpha+\alpha}*\lim_{{u} \to {0}} \frac {\sin{u}}u =\lim_{{x} \to {\alpha}} \frac {\sin{2\alpha} } {2\alpha}*\lim_{{u} \to {0}} \frac {\sin{u}}u

Primeniti tablični limes: lim⁡x→0sin⁡xx=1 \lim_{{x} \to {0}}\frac {\sin{x}} {x}=1

lim⁡x→αsin⁡2α2α∗1=lim⁡x→αsin⁡2α2α\lim_{{x} \to {\alpha}} \frac {\sin{2\alpha} } {2\alpha}*1=\lim_{{x} \to {\alpha}} \frac {\sin{2\alpha} } {2\alpha}

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