Konsultime Mat avancuar 26.06.2013. Limiti 1

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– Posted on 26/06/2013Posted in: USHTRIME MATEMATIKE

Gjeni limitet:  

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

b) \lim_{x\rightarrow 0}\frac{cos3x-cosx}{cosx-1}  

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

d)\lim_{x\rightarrow \frac{\pi }{2}}\frac{\pi -2x}{cosx}   

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

zgjidhje

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

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

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

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

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

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

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