Efectueaza o cautare in website!

Informaţii, definiţii, teoreme, formule, exerciţii şi probleme rezolvate din matematica de liceu.

Data publicarii: 04 Noiembrie, 2014

EXERCITIUL 9

Suport teoretic:

Functii trigonometrice,primitive,functii continue,regula lui l'Hospital.

Enunt:

Sa se determine m real, astfel incat functia

f:[-\frac{\pi}{2a},\frac{\pi}{2a}]\rightarrow{R},f:[-\frac{\pi}{2a},\frac{\pi}{2a}]\rightarrow{R}, f(x)=\begin{cases}{{(cosx)}^{ctg(ax)}},\,{x}\in{[-\frac{\pi}{2a},0)}\\m,\,x=0\\{{(ctgx)}^{sin(ax)}},\,x\in{(0,\frac{\pi}{2a}]}\end{cases},f(x)=\begin{cases}{{(cosx)}^{ctg(ax)}},\,{x}\in{[-\frac{\pi}{2a},0)}\\m,\,x=0\\{{(ctgx)}^{sin(ax)}},\,x\in{(0,\frac{\pi}{2a}]}\end{cases},

unde a > 1, sa admita primitive.

Raspuns: 

m = 1.

Rezolvare:

Se calculeaza m astfel incat functia f sa fie continua in x = 0:

\lim_{{x}{\nearrow}{0}}f(x)=\lim_{{x}{\searrow}{0}}f(x)=m.\lim_{{x}{\nearrow}{0}}f(x)=\lim_{{x}{\searrow}{0}}f(x)=m.

Obtinem succesiv:

\lim_{{x}{\nearrow}{0}}f(x)=\lim_{{x}{\nearrow}{0}}{(cosx)}^{ctg(ax)}=(1^{\infty})=\lim_{{x}{\nearrow}{0}}f(x)=\lim_{{x}{\nearrow}{0}}{(cosx)}^{ctg(ax)}=(1^{\infty})= {e}^{{\lim_{x}{\nearrow}{0}}{[ctg(ax)]\cdot[\ln{(cosx)}]}}={e}^{{\lim_{x}{\nearrow}{0}}{[ctg(ax)]\cdot[\ln{(cosx)}]}}=

={e}^{\lim_{x}{\nearrow}{0}{\frac{ln(cosx)}{tg(ax)}}}=(\frac{0}{0};{L^{={e}^{\lim_{x}{\nearrow}{0}{\frac{ln(cosx)}{tg(ax)}}}=(\frac{0}{0};{L^{'}}{Hospital})=\cdots= e^0=1

si 

\lim_{{x}{\searrow}{0}}f(x) = \lim_{x\searrow{0}}{{(ctgx)}^{sin(ax)}}= ({\infty}^{0})=\lim_{{x}{\searrow}{0}}f(x) = \lim_{x\searrow{0}}{{(ctgx)}^{sin(ax)}}= ({\infty}^{0})= {e}^{{\lim_{x\searrow{0}}{[sin(ax)}]\cdot[{ln(ctgx)}]}}= ({0}\cdot{\infty})={e}^{{\lim_{x\searrow{0}}{[sin(ax)}]\cdot[{ln(ctgx)}]}}= ({0}\cdot{\infty})=

=e^{\lim_{x\searrow{0}}\frac{ln(ctgx)}{\frac{1}{sin(ax)}}}==e^{\lim_{x\searrow{0}}\frac{ln(ctgx)}{\frac{1}{sin(ax)}}}= (\frac{\infty}{\infty};{L^{(\frac{\infty}{\infty};{L^{'}}{Hospital})=\cdots=1.

Deci f este continua in x = 0, daca m = 1.


Adăugaţi un comentariu

Adăugaţi un comentariu
Introdu codul din imagine

Răspunsuri şi comentarii

Până acum, niciun comentariu nu a fost adăugat.

 

CATEGORII :


Arhiva blog-ului

 

 
Developed by Hagau Ioan