Trigonometrijski limes: sin x kad x teži pi

TEKST ZADATKA

Odrediti graničnu vrednost:

\lim_{{x} \to {\pi}} \frac {\sin{x}} {1-\frac {x^2} {\pi^2}}

REŠENJE ZADATKA

Srediti imenilac:

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

Osloboditi se dvojnog razlomka:

\lim_{{x} \to {\pi}} \frac {\pi^2*\sin{x}} {\pi^2-x^2}

Primeniti formulu za razliku kvadrata:

\lim_{{x} \to {\pi}} \frac {\pi^2*\sin{x}} {(\pi-x)(\pi+x)}

Izvući konstante ispred limesa:

\pi^2*\lim_{{x} \to {\pi}} \frac {\sin{x}} {(\pi-x)(\pi+x)}

Uvesti smenu:

\pi-x=u \implies x=\pi-x

Uvrstiti smenu u izraz:

\pi^2*\lim_{{u} \to {0}} \frac {\sin{(\pi-u)}} {u(\pi+\pi-u)}= \pi^2*\lim_{{u} \to {0}} \frac {\sin{(\pi-u)}} {u(2\pi-u)}

Primeniti adicionu formulu za sinus razlike: $\sin{(\alpha-\beta)}=\sin{\alpha}\cos{\beta}-\sin{\beta}\cos{\alpha}$

\pi^2*\lim_{{u} \to {0}} \frac {\sin{\pi}\cos{u}-\sin{u}\cos{\pi}} {u(2\pi-u)}

Uvrstiti vrednosti trigonometrijskih funkcija:

\pi^2*\lim_{{u} \to {0}} \frac {0*\cos{u}-\sin{u}*(-1)} {u(2\pi-u)}= \pi^2*\lim_{{u} \to {0}} \frac {\sin{u}} {u(2\pi-u)}

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

\pi^2* \frac 1 {2\pi-u}

Zameniti $u=0.$

\pi^2* \frac 1 {2\pi-0}=\pi^2* \frac 1 {2\pi}=\frac {\pi} 2

7. Trigonometrijski limes