3 granice

[Hania_87]

1) \(\displaystyle{ \lim_{x\to 0} \frac {\arcsin 2x-2 arcsinx}{x^{3}}}\)
2) \(\displaystyle{ \lim_{x\to } \sqrt{x+3} \sin (\sqrt{x+2}-\sqrt{x+1})}\)
3) \(\displaystyle{ \lim_{x\to \frac {\pi}{2}}e^{cos x}}\)

[Kostek]

2.
\(\displaystyle{ \lim_{x\to{\infty}}\sqrt{x+3}sin(\frac{1}{\sqrt{x+2}+\sqrt{x+1}})=\lim_{x\to{\infty}}\frac{\sqrt{x+3}}{\sqrt{x+2}+\sqrt{x+1}}\frac{sin(\frac{1}{\sqrt{x+2}+\sqrt{x+1}})}{\frac{1}{\sqrt{x+2}+\sqrt{x+1}}}{}}=\frac{1}{2}}\)

3
\(\displaystyle{ \lim_{x\to{\frac{\pi}{2}}}e^{cosx}=e^{0}=1}\)

[max]

1) Korzystając z reguły de L'Hospitala:
\(\displaystyle{ \lim_{x\to 0} \frac{\arcsin 2x - 2\arcsin x}{x^{3}} \stackrel{\mathbf{H}}{=} \lim_{x\to 0} \frac{(\arcsin 2x - 2\arcsin x)'}{(x^{3})'} = \\
= \lim_{x\to 0}\frac{\frac{2}{\sqrt{1 - 4x^{2}}} - \frac{2}{\sqrt{1 - x^{2}}}}{3x^{2}} = \lim_{x\to 0}\frac{2(\sqrt{1 - x^{2}} -\sqrt{1 - 4x^{2}})}{3x^{2}\sqrt{(1 - 4x^{2})(1 - x^{2})}} =\\
= \lim_{x\to 0}\frac{2(\sqrt{1 - x^{2}} -\sqrt{1 - 4x^{2}})}{3x^{2}\sqrt{(1 - 4x^{2})(1 - x^{2})}}\cdot \frac{\sqrt{1 - x^{2}} +\sqrt{1 - 4x^{2}}}{ \sqrt{1 - x^{2}} +\sqrt{1 - 4x^{2}}}=\\
= \lim_{x\to 0}\frac{2\cdot 3x^{2}}{3x^{2}(\sqrt{1 - x^{2}} +\sqrt{1 - 4x^{2}})\sqrt{(1 - 4x^{2})(1 - x^{2})}} = \\
= \lim_{x\to 0}\frac{2}{(\sqrt{1 - x^{2}} +\sqrt{1 - 4x^{2}})\sqrt{(1 - 4x^{2})(1 - x^{2})}}= 1}\)