Limiti i funksionit Matura 2013. Forma e pacaktuar ∞-∞,∞/∞,0/0

By
– Posted on 27/06/2013Posted in: USHTRIME MATEMATIKE

Gjeni limitet:

a)\lim_{x\rightarrow -\infty }\left ( \sqrt{x^{2}+1}+x \right ), 

b) \lim_{x\rightarrow 0}\frac{\sqrt{x^{2}+1}-1}{\sqrt{x^{2}+16}-4} , 

c) \lim_{x\rightarrow +\infty }\frac{\sqrt{x^{2}+1}+\sqrt{x}}{\sqrt[4]{x^{3}+x}-x}

zgjidhje

a) \lim_{x\rightarrow -\infty }\left ( \sqrt{x^{2}+1}+x \right )=\lim_{x\rightarrow -\infty }\frac{\left ( \sqrt{x^{2}+1}+x \right )\left ( \sqrt{x^{2}+1}-x \right )}{\left ( \sqrt{x^{2}+1}-x \right )}=\lim_{x\rightarrow -\infty }\frac{x^{2}+1-x^{2}}{\sqrt{x^{2}+1}-x}=\lim_{x\rightarrow -\infty }\frac{1}{\sqrt{x^{2}+1}-x}=0  sepse \lim_{x\rightarrow -\infty }\left ( \sqrt{x^{2}+1}-x \right )=\lim_{x\rightarrow -\infty }\left ( \sqrt{x^{2}\left ( 1+\frac{1}{x^{2}} \right )}-x \right )=\lim_{x\rightarrow -\infty }\left [ -x\left ( \sqrt{1+\frac{1}{x^{2}}}+1 \right ) \right ] =+\infty       \sqrt{x^{2}}=-x  per x<0

b)\lim_{x\rightarrow 0}\frac{\sqrt{x^{2}+1}-1}{\sqrt{x^{2}+16}-4}=\lim_{x\rightarrow 0}\frac{\left ( \sqrt{x^{2}+1}-1 \right )\left ( \sqrt{x^{2}+1}+1 \right )\left (\sqrt{x^{2}+16}+4 \right )}{\left ( \sqrt{x^{2}+16}-4 \right )\left ( \sqrt{x^{2}+16}+4 \right )\left ( \sqrt{x^{2}+1}+1 \right )}=\lim_{x\rightarrow 0}\frac{x^{2}\left ( \sqrt{x^{2}+16}+4 \right )}{x^{2}\left ( \sqrt{x^{2}+1}+1 \right )}=\frac{8}{4}=4

c) \lim_{x\rightarrow +\infty }\frac{\sqrt{x^{2}+1}+\sqrt{x}}{\sqrt[4]{x^{3}+x}-x}=\lim_{x\rightarrow +\infty }\frac{x\left ( \sqrt{1+\frac{1}{x^{2}}}+\sqrt{\frac{1}{x}} \right )}{x\left ( \sqrt[4]{\frac{1}{x}+\frac{1}{x^{3}}}-1 \right )}=-1  Kryej veprimet qe nga faktorizimi brenda rrenjeve, pastaj thjeshto me x .

Join the Forum discussion on this post

Related Posts