by Tìm các giới hạn sau:
a) $\mathop {\lim }\limits_{x \to + \infty } {e^{\frac{{2x + 1}}{{x + 1}}}}.$
b) $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{{e^x}}} – 1}}{{2x}}.$
Lời giải
a) $\mathop {\lim }\limits_{x \to + \infty } {e^{\frac{{2x + 1}}{{x + 1}}}} = {e^{\mathop {\lim }\limits_{x \to + \infty } \frac{{2x + 1}}{{x + 1}}}} = {e^2}.$
b) $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{{e^x}}} – 1}}{{2x}} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^{\frac{x}{3}}} – 1}}{{2x}}$ $ = \mathop {\lim }\limits_{x \to 0} \left( {\frac{1}{6}.\frac{{{e^{\frac{x}{3}}} – 1}}{{\frac{x}{3}}}} \right) = \frac{1}{6}.$
