Trigonometrijski limes: razlika kvadrata sinusa

TEKST ZADATKA

Odrediti graničnu vrednost:

\lim_{{x} \to {\alpha}} \frac {\sin^2{x}-\sin^2{\alpha}} {x^2-{\alpha}^2}

REŠENJE ZADATKA

Primeniti formulu za sinus poluugla: $|\sin{\frac {\alpha} 2}|=\sqrt{\frac {1-\cos{\alpha}} 2}$

\lim_{{x} \to {\alpha}} \frac {\frac {1-\cos{2x}} 2-\frac {1-\cos{2\alpha}} 2} {x^2-{\alpha}^2}

Osloboditi se dvojnog razlomka:

\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} \to {\alpha}} \frac {{\cos{2\alpha}-\cos{2x}} } {2(x-\alpha)(x+\alpha)}

Primeniti formulu za transformaciju razlike kosinusa: $\cos{\alpha}-\cos{\beta}=-2\sin{\frac {\alpha+\beta} 2}\sin{\frac {\alpha-\beta} 2}$

\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} \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} \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} \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} \to {\alpha}} \frac {\sin{(\alpha+x)} } {\alpha+x}*\lim_{{x} \to {\alpha}} \frac {\sin{(x-\alpha)}} {x-\alpha}

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

\frac {\sin{(\alpha+\alpha)} } {\alpha+\alpha}* \frac {\sin{(\alpha-\alpha)}} {\alpha-\alpha}=\frac 0 0

Uvesti smenu:

x-\alpha=u

Uvrstiti smenu u izraz:

\lim_{{x} \to {\alpha}} \frac {\sin{(\alpha+x)} } {\alpha+x}*\lim_{{u} \to {0}} \frac {\sin{u}}u

Ponovo pokušati zameniti $x=\alpha.$

\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: $\lim_{{x} \to {0}}\frac {\sin{x}} {x}=1$

\lim_{{x} \to {\alpha}} \frac {\sin{2\alpha} } {2\alpha}*1=\lim_{{x} \to {\alpha}} \frac {\sin{2\alpha} } {2\alpha}

5. Trigonometrijski limes