problem1
\[\text { 求极限: } \lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{\ln (1+\sin x)} \text {. } \]
solution1
\[\begin{aligned} &\text { 解: } \lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{\ln (1+\sin x)}=\lim _{x \rightarrow 0} \frac{e^{x}-1+1-e^{-x}}{\sin x} \\ &=\lim _{x \rightarrow 0} \frac{e^{x}-1-\left(e^{-x}-1\right)}{x}=\lim _{x \rightarrow 0} \frac{x-(-x)}{x}=2 \end{aligned}\]
problem2
\[\text { 求极限: } \lim _{x \rightarrow 0} \frac{\sin x-x \cos x}{e^{\ln \left(1+\tan ^{3} x\right)}-1} \text {. } \]
solution2
\[ \begin{aligned} &\text { 解: } \lim _{x \rightarrow 0} \frac{\sin x-x \cos x}{e^{\ln \left(1+\tan ^{3} x\right)}-1}=\lim _{x \rightarrow 0} \frac{\sin x-x \cos x}{\ln \left(1+\tan ^{3} x\right)}=\lim _{x \rightarrow 0} \frac{\sin x-x \cos x}{\tan ^{3} x} \\ &=\lim _{x \rightarrow 0} \frac{\sin x-x \cos x}{x^{3}}=\lim _{x \rightarrow 0} \frac{\sin x-x+x-x \cos x}{x^{3}} \\ &=\lim _{x \rightarrow 0} \frac{\sin x-x}{x^{3}}+\lim _{x \rightarrow 0} \frac{x(1-\cos x)}{x^{3}}=\lim _{x \rightarrow 0} \frac{-\frac{1}{6} x^{3}}{x^{3}}+\lim _{x \rightarrow 0} \frac{x \cdot \frac{1}{2} x^{2}}{x^{3}} \\ &=-\frac{1}{6}+\frac{1}{2}=\frac{1}{3} . \end{aligned} \]
problem3
\[\text { 求极限: } \lim _{x \rightarrow 0} \frac{\int_{0}^{x}\left[(3+2 \tan t)^{t}-3^{t}\right] d t}{e^{3 x^{3}}-1} \text {. } \]
solution3
\[ \begin{aligned} &\text { 解: } \lim _{x \rightarrow 0} \frac{\int_{0}^{x}\left[(3+2 \tan t)^{t}-3^{t}\right] d t}{e^{3 x^{3}}-1}=\lim _{x \rightarrow 0} \frac{\int_{0}^{x}\left[(3+2 \tan t)^{t}-3^{t}\right] d t}{3 x^{3}} \\ &=\lim _{x \rightarrow 0} \frac{(3+2 \tan x)^{x}-3^{x}}{9 x^{2}}=\lim _{x \rightarrow 0} \frac{e^{x \ln (3+2 \tan x)}-e^{x \ln 3}}{9 x^{2}} \\ &=\lim _{x \rightarrow 0} \frac{x \ln (3+2 \tan x)-x \ln 3}{9 x^{2}}=\lim _{x \rightarrow 0} \frac{\ln (3+2 \tan x)-\ln 3}{9 x} \\ &=\lim _{x \rightarrow 0} \frac{\ln \left(\frac{3+2 \tan x}{3}\right)}{9 x}=\lim _{x \rightarrow 0} \frac{\ln \left(1+\frac{2}{3} \tan x\right)}{9 x}=\lim _{x \rightarrow 0} \frac{\frac{2}{3} \tan x}{9 x} \\ &=\frac{2}{27} \lim _{x \rightarrow 0} \frac{x}{x}=\frac{2}{27} . \end{aligned}\]
problem4
\[\text { 求极限: } \lim _{x \rightarrow 0} \frac{\sqrt{1+3 x}-\sqrt[3]{1+2 x}}{\ln (1+x)} \text {. } \]
solution4
\[\begin{aligned} &\text { 解: } \lim _{x \rightarrow 0} \frac{\sqrt{1+3 x}-\sqrt[3]{1+2 x}}{\ln (1+x)}=\lim _{x \rightarrow 0} \frac{\sqrt{1+3 x}-\sqrt[3]{1+2 x}}{x} \\ &=\lim _{x \rightarrow 0} \frac{\sqrt{1+3 x}-1+1-\sqrt[3]{1+2 x}}{x}=\lim _{x \rightarrow 0} \frac{\sqrt{1+3 x}-1-(\sqrt[3]{1+2 x}-1)}{x} \\ &=\lim _{x \rightarrow 0} \frac{\sqrt{1+3 x}-1}{x}-\lim _{x \rightarrow 0} \frac{\sqrt[3]{1+2 x}-1}{x}=\lim _{x \rightarrow 0} \frac{(1+3 x)^{\frac{1}{2}}-1}{x}-\lim _{x \rightarrow 0} \frac{(1+2 x)^{\frac{1}{3}}-1}{x} \\ &=\lim _{x \rightarrow 0} \frac{\frac{1}{2} \cdot 3 x}{x}-\lim _{x \rightarrow 0} \frac{\frac{1}{3} \cdot 2 x}{x}=\frac{3}{2}-\frac{2}{3}=\frac{5}{6} . \end{aligned} \]