Limits

$$ \begin{aligned} \text{Tính: }& L = \lim_{x \to +\infty}\bigg(\frac{x+1}{x+3}\bigg)^{x}\\ \text{Giải: } & \\ \text{Ta có: }& \\ L&=\lim_{x \to +\infty}\bigg(\frac{x+1}{x+3}\bigg)^x \\ &= \lim_{x \to +\infty}\bigg(\frac{x+3-2}{x+3}\bigg)^x \\ &= \lim_{x \to +\infty}\bigg(1 + \frac{-2}{x+3}\bigg)^x \\ \text{Đặt: } u&=\frac{-2}{x+3} \\ &\Leftrightarrow x+3 = \frac{-2}{u} \\ &\Leftrightarrow x = \frac{-2}{u} - 3 \\ \text{Vì: } &\lim_{x \to +\infty}\bigg(\frac{-2}{x+3}\bigg) = 0 \Rightarrow \lim_{x \to +\infty}u = \lim_{u \to 0}u = 0 \\ \text{Do đó: } &\\ L &= \lim_{x \to +\infty}\bigg(1 + \frac{-2}{x+3}\bigg)^x \\ &= \lim_{u \to 0}\bigg(1+u\bigg)^{\frac{-2}{u} -3} \\ &= \lim_{u \to 0}\bigg(1+u\bigg)^{\frac{-2}{u}}.\bigg(1+u\bigg)^{-3} \\ &= \lim_{u \to 0}[\bigg(1+u\bigg)^{\frac{1}{u}}]^{-2}.\lim_{u \to 0}\bigg(1+u\bigg)^{-3} \\ &= \lim_{u \to 0}e^{\lim_{u \to 0}-2}.\lim_{u \to 0}1^{\lim_{u \to 0}-3} \\ &= e^{-2}.1^{-3} \\ &= e^{-2} \\ &\text{Vậy } L = e^{-2} \square \end{aligned} $$

$$ \begin{aligned} \text{Tính: }& L = \lim_{x \to 0}{\frac{\sqrt{1 + 4x} - 1}{\ln{1 + 3x}}} \\ \text{Giải: }& \\ L=&\lim_{x \to 0}{\frac{\sqrt{1 + 4x} - 1}{\ln{(1 + 3x)}}} \\ =&\lim_{x \to 0}{\frac{(\sqrt{1 + 4x} - 1)(\sqrt{1 + 4x} + 1)}{\ln{(1 + 3x)}(\sqrt{1 + 4x} + 1)}} \\ =&\lim_{x \to 0}{\frac{\sqrt{1 + 4x}^{2} - 1^{2}}{\ln{(1 + 3x)}(\sqrt{1 + 4x} + 1)}} \\ =&\lim_{x \to 0}{\frac{4x}{\ln{(1 + 3x)}(\sqrt{1 + 4x} + 1)}} \\ =&\lim_{x \to 0}{\frac{4x}{\ln{(1 + 3x)}}} \cdot \lim_{x \to 0}{\frac{1}{\sqrt{1 + 4x} + 1}} \\ =&\lim_{x \to 0}{\bigg(\frac{4}{3}\cdot\frac{3x}{\ln{(1 + 3x)}}\bigg)} \cdot \lim_{x \to 0}{\frac{1}{\sqrt{1 + 4x} + 1}} \\ =&\lim_{x \to 0}{\bigg(\frac{4}{3}\cdot 1 \bigg)} \cdot \lim_{x \to 0}{\frac{1}{\sqrt{1 + 4x} + 1}} \\ =&\bigg(\frac{4}{3}\cdot 1 \bigg) \cdot \frac{1}{\sqrt{1 + 4 \cdot 0} + 1} \\ =&\frac{4}{3} \cdot \frac{1}{2} \\ =&\frac{2}{3} \square\\ \end{aligned} $$