Limite L`Hospital
Re: Limite L`Hospital
Daca vrei tu poti aplica regula lui L`Hospital, eu o sa aplic niste limite remarcabile:
h)-sin^{2}x}{-\frac{\left ( e^{-x^{2}}-1 \right )}{-x^{2}}\cdot \left ( -x^{2} \right )}=\lim_{x\rightarrow 0}\frac{ln^{2}\left ( x+1 \right )-sin^{2}x}{x^{2}}=\lim_{x\rightarrow 0}\left ( \frac{ln\left ( x+1 \right )}{x} \right )^{2}-\left ( \frac{sinx}{x} \right )^{2}=1-1=0)
i)
j)\cdot \left ( \sqrt{tg^{2}x}+\sqrt{tgxsinx}+\sqrt{sin^{2}x} \right )}{x^{3}\cdot \sqrt{x}}=\lim_{x\rightarrow 0}\frac{\sqrt{tgx}-\sqrt{sinx}}{\sqrt{x}}\cdot \frac{\sqrt{sin^{2}x}}{x^{3}}\cdot \left ( \frac{1}{\sqrt{cos^{2}x}}+\frac{1}{\sqrt{cosx}}+1 \right )=3\cdot\lim_{x\rightarrow 0}\frac{\sqrt{tgx}-\sqrt{sinx}}{\sqrt{x}}\cdot \sqrt{\left ( \frac{sinx}{x} \right )^{2}}\cdot \frac{1}{x^{2}}=3\cdot \lim_{x\rightarrow 0}\sqrt{\frac{sinx}{x}}\cdot \left ( \frac{1}{\sqrt{cosx}}-1 \right )\cdot \frac{1}{x^{2}}=3\lim_{x\rightarrow 0}\frac{\frac{1}{\sqrt{cosx}}-1}{x^{2}})
Aici aplicam regula lui l'Hospital si obtinem :
Daca am gresit la calcule ,te rog sa ma corectezi!
h)
i)
j)
Aici aplicam regula lui l'Hospital si obtinem :
Daca am gresit la calcule ,te rog sa ma corectezi!