# Konsultime Mat avancuar 26.06.2013. Limiti 1

Gjeni limitet:

a) $\lim_{x\rightarrow&space;0}\frac{2cosx\left&space;(&space;1-cosx&space;\right&space;)}{sin^{2}x}$

b) $\lim_{x\rightarrow&space;0}\frac{cos3x-cosx}{cosx-1}$

c)$\lim_{x\rightarrow&space;0}\frac{1-cos^{2}x}{x\cdot&space;sinx}$

d)$\lim_{x\rightarrow&space;\frac{\pi&space;}{2}}\frac{\pi&space;-2x}{cosx}$

e)$\lim_{x\rightarrow&space;0}\frac{\sqrt{1+xsinx}-\sqrt{cos2x}}{tg^{2}x}$

zgjidhje

a) $\lim_{x\rightarrow&space;0}\frac{2cosx\left&space;(&space;1-cosx&space;\right&space;)}{sin^{2}x}=\lim_{x\rightarrow&space;0}\left&space;[\frac{1-cosx}{x^{2}}\cdot&space;\frac{x^{2}}{sin^{2}x}\cdot&space;2cosx&space;\right&space;]=\frac{1}{2}\cdot&space;1\cdot&space;2=1$

b) $\lim_{x\rightarrow&space;0}\frac{cos3x-cosx}{cosx-1}=\lim_{x\rightarrow&space;0}\frac{\left&space;(1-cosx&space;\right&space;)-\left&space;(&space;1-cos3x&space;\right&space;)}{-\left&space;(&space;1-cosx&space;\right&space;)}=\lim_{x\rightarrow&space;0}\left&space;[&space;\frac{1-cos3x}{1-cosx}-1&space;\right&space;]=\lim_{x\rightarrow&space;0}\left&space;[\frac{1-cos3x}{(3x)^{2}}\cdot&space;\frac{x^{2}}{1-cosx}\cdot&space;9-1&space;\right&space;]=\frac{1}{2}\cdot&space;2\cdot&space;9-1=8$

c) $\lim_{x\rightarrow&space;0}\frac{1-cos^{2}x}{x\cdot&space;sinx}=\lim_{x\rightarrow&space;0}\left&space;[\frac{1-cosx}{x^{2}}\cdot\frac{x}{sinx}&space;\right&space;]=\frac{1}{2}\cdot&space;1=\frac{1}{2}$

d) Zevendesojme  $\frac{\pi&space;}{2}-x=t$$x\rightarrow&space;\frac{\pi&space;}{2}\Rightarrow&space;t\rightarrow&space;0$  atehere$\lim_{x\rightarrow&space;\frac{\pi&space;}{2}}\frac{\pi&space;-2x}{cosx}=\lim_{t\rightarrow&space;0}\frac{\pi&space;-2\left&space;(&space;\frac{\pi&space;}{2}-t&space;\right&space;)}{cos\left&space;(&space;\frac{\pi&space;}{2}-t&space;\right&space;)}=\lim_{t\rightarrow&space;0}\frac{2t}{sint}=2\cdot&space;\lim_{t\rightarrow&space;0}\frac{t}{sint}=2$

e)$\lim_{x\rightarrow&space;0}\frac{\sqrt{1+xsinx}-\sqrt{cos2x}}{tg^{2}x}=\lim_{x\rightarrow&space;0}\frac{\left&space;(\sqrt{1+xsinx}-\sqrt{cos2x}&space;\right&space;)\cdot&space;\left&space;(&space;\sqrt{1+xsinx}+\sqrt{cos2x}&space;\right&space;)}{tg^{2}x\cdot&space;\left&space;(&space;\sqrt{1+xsinx}+\sqrt{cos2x}&space;\right&space;)}=\lim_{x\rightarrow&space;0}\left&space;[\frac{1+xsinx-cos2x}{tg^{2}x\cdot&space;\left&space;(&space;\sqrt{1+xsinx}+\sqrt{cos2x}&space;\right&space;)}&space;\right&space;]=\lim_{x\rightarrow&space;0}\left&space;[&space;\frac{1-cos2x+xsinx}{tg^{2}x}&space;\right\cdot&space;\frac{1}{\sqrt{1+xsinx}+\sqrt{cos2x}}&space;]=\lim_{x\rightarrow&space;0}\left&space;[&space;\left&space;(\frac{1-cos2x}{tg^{2}x}+\frac{xsinx}{tg^{2}x}&space;\right&space;)\cdot&space;\left&space;(&space;\frac{1}{\sqrt{1+xsinx}+\sqrt{cos2x}}&space;\right&space;)&space;\right&space;]=....$

Rregullo vetem faktorin  e pare ne kllapa $\frac{1-cos2x}{\left&space;(&space;2x&space;\right&space;)^{2}}\cdot&space;4\cdot&space;\frac{x^{2}}{tg^{2}x}+\frac{x}{sinx}\cdot&space;cos^{2}x$   dhe llogarit limitin.

Problemi eshte te veqosh limitet e rendesishme te:   $\frac{sinax}{ax}$  , $\frac{1-cos\left&space;(ax&space;\right&space;)}{(ax)^{2}}$ , $\frac{tg(ax)}{ax}$  etj. sepse ne shumicen e rasteve tek ato eshte forma e pacaktuar