Tính giới hạn:
[tex]\lim_{x\rightarrow 0}\frac{e^{2x}-\sqrt{2x+1}}{\sqrt{3x+4}-2-x}[/tex]
mình giải theo cách này nhé:
[tex]\lim_{x->0}\frac{(e^{4x}-2x-1)(\sqrt{3x+4}+2+x)}{(-x^{2}-x)(e^{2x}+\sqrt{2x+1})}[/tex]
[tex]=\lim_{x->0}\frac{(e^{4x}-1)(\sqrt{3x+4}+2+x)}{4x(\frac{-x}{4}-\frac{1}{4})(e^{2x}+\sqrt{2x+1}}+\lim_{x->0}\frac{(-2x)(\sqrt{3x+4}+2+x)}{x(-x-1)(e^{2x}+\sqrt{2x+1})}[/tex]
[tex]=\lim_{x->0}\frac{(e^{4x}-1)}{4x}.\lim_{x->0}\frac{(\sqrt{3x+4}+2+x)}{(\frac{-x}{4}-\frac{1}{4})(e^{2x}+\sqrt{2x+1})}+\lim_{x->0}\frac{2(\sqrt{3x+4}+2+x)}{(x+1)(e^{2x}+\sqrt{2x+1})}[/tex]
[tex]=1.\lim_{x->0}\frac{(\sqrt{3x+4}+2+x)}{(\frac{-x}{4}-\frac{1}{4})(e^{2x}+\sqrt{2x+1})}+\lim_{x->0}\frac{2(\sqrt{3x+4}+2+x)}{(x+1)(e^{2x}+\sqrt{2x+1})}[/tex]
[tex]=-4[/tex]