Намерете

[Гост]

[tex]\lim_{x \to 0}[/tex] [tex]\frac{sin4x}{ \sqrt{x+1}-1 }[/tex]

[nikola.topalov]

$$\lim\limits_{x\to 0}\dfrac{\sin 4x}{\sqrt{x+1}-1}=\lim\limits_{x\to 0}\dfrac{(\sqrt{x+1}+1)\sin 4x}{x}=4\lim\limits_{\substack{x\to 0 \\ 4x\to 0}} \dfrac{\sin 4x}{4x}\lim\limits_{x\to 0}(\sqrt{x+1}+1)=8$$