Sa se afle limitele:
a)
![\[ {\lim }\limits_{x \to \frac{\pi }{3}} \frac{{tg^3 x - 3tgx}}{{\cos (x + \frac{\pi }{6})}} \]](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-55acb5b41bbc17384905236b9ab71e6f_l3.png "Rendered by QuickLaTeX.com")
b)
![\[ {\lim }\limits_{x \to 0} \frac{{tg(a + x)tg(a - x) - tg^2 a}}{{x^2 }} \]](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-a250476c400863d9cab2d14d19aa269f_l3.png "Rendered by QuickLaTeX.com")
c)
![\[ {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{1 - ctg^3 x}}{{2 - ctgx - ctg^3 x}} \]](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-9d2b84510b703437b00d3601b7aba8f7_l3.png "Rendered by QuickLaTeX.com")
d)
![\[ {\lim }\limits_{x \to - \infty } \frac{{\ln (1 + 3^x )}}{{\ln (1 + 2^x )}} \]](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-f008c77b23dd130783b63ae79893bdb5_l3.png "Rendered by QuickLaTeX.com")
AniDeȘcoală.ro
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Please briefly explain why you feel this question should be reported.
Va rugam explicate, pe scurt, de ce credeti ca aceasta intrebare trebuie raportata.
Motivul pentru care raportezi utilizatorul.
1)Sa prelucram mai intai expresia de sub limita ;E=tgx.((tgx)^2-(tg(pi/3))^2)/
cos(x+(pi/6))=tgX.(tgx+tg(pi/3))(tgx-tg(pi/3)) / cos(x+(pi/6))=tgx.sin(x+(pi/3)).sin(x-(pi/3)) / sin((pi/2)-x-(pi/6)),(cosx)^2.(cos(pi/3))^2=-tgx.sin(x+(pi/3)) / (cosx)^2.(cos(pi/3))^2. Cand x->(pi/3) expresia E->(sqrt(3)).(sqrt(3))/2 / (1/2)^2.(1/2)^2=3/8
2)Sa prelucram si aceasta expresie ; E=(1/x^2).{[(tga+tgx)/(1-tga,tgx)][(tga-
tgx0/(1+tga.tgx)]-(tga)^2}=(1/x^2).{[(tga)^2-(tgx)^2]/[1-(tga)^2.(tgx)^2]
-(tga)^2}=(1/x^2).(tgx)^2,[(tga)^4-1]/[1-(tga)^2.(tgx)^2]. cand x-.>0
E->1.[(tga)^4-1]/1=(tga)^4-1.
3)Si in acest caz prelucram expresia ;
E=(1-ctgx)(1+ctgx+(ctgx)^2)/[(1-ctgx)(2+ctgx+(ctgx)^2)]=(1+ctgx+(ctgx)^2)/(2+ctgx+(ctgx)^2). cand x->(pi/4) atunci E->3/4 (ctg(pi/4)=1)
4)Prelucrand expresia data avem; {ln[(3^x.(1+1/3^x)]}/{ln[(2^x.(1+1/2^x)]}={x.ln3+ln(1+1/3^x)}/{x.ln2+ln(1+1/2^x)} Cand x->+infinit
E->ln3/ln2