192.

Trigonometrijski limes

TEKST ZADATKA

Odrediti graničnu vrednost:

lim⁡x→πsin⁡x1−x2π2\lim_{{x} \to {\pi}} \frac {\sin{x}} {1-\frac {x^2} {\pi^2}}
REŠENJE ZADATKA

Srediti imenilac:

lim⁡x→πsin⁡xπ2−x2π2\lim_{{x} \to {\pi}} \frac {\sin{x}} {\frac {\pi^2-x^2} {\pi^2}}

Osloboditi se dvojnog razlomka:

lim⁡x→ππ2∗sin⁡xπ2−x2\lim_{{x} \to {\pi}} \frac {\pi^2*\sin{x}} {\pi^2-x^2}

Primeniti formulu za razliku kvadrata:

lim⁡x→ππ2∗sin⁡x(π−x)(π+x)\lim_{{x} \to {\pi}} \frac {\pi^2*\sin{x}} {(\pi-x)(\pi+x)}

Izvući konstante ispred limesa:

π2∗lim⁡x→πsin⁡x(π−x)(π+x)\pi^2*\lim_{{x} \to {\pi}} \frac {\sin{x}} {(\pi-x)(\pi+x)}

Uvesti smenu:

π−x=u  ⟹  x=π−x\pi-x=u \implies x=\pi-x
DODATNO OBJAŠNJENJE

Uvrstiti smenu u izraz:

π2∗lim⁡u→0sin⁡(π−u)u(π+π−u)=π2∗lim⁡u→0sin⁡(π−u)u(2π−u)\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⁡(α−β)=sin⁡αcos⁡β−sin⁡βcos⁡α\sin{(\alpha-\beta)}=\sin{\alpha}\cos{\beta}-\sin{\beta}\cos{\alpha}

π2∗lim⁡u→0sin⁡πcos⁡u−sin⁡ucos⁡πu(2π−u) \pi^2*\lim_{{u} \to {0}} \frac {\sin{\pi}\cos{u}-\sin{u}\cos{\pi}} {u(2\pi-u)}

Uvrstiti vrednosti trigonometrijskih funkcija:

π2∗lim⁡u→00∗cos⁡u−sin⁡u∗(−1)u(2π−u)=π2∗lim⁡u→0sin⁡uu(2π−u) \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)}
DODATNO OBJAŠNJENJE

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

π2∗12π−u\pi^2* \frac 1 {2\pi-u}
DODATNO OBJAŠNJENJE

Zameniti u=0.u=0.

π2∗12π−0=π2∗12π=π2\pi^2* \frac 1 {2\pi-0}=\pi^2* \frac 1 {2\pi}=\frac {\pi} 2

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