Template:Sandbox：修订间差异

←上一编辑下一编辑→

@@ 第1行： / 第1行： @@
 {{SandboxNote}}
+<math>
+(2)
+解：\lim_{x\rightarrow 0}f(x)=\lim_{x\rightarrow 0}\sin x\cos \frac{1}{x}=0
+\\
+所以f(0)=0
+\\
+(3)
+解：若k=0，则\lim_{x\rightarrow 0}f(x)=\lim_{x\rightarrow 0}\ln 1^{\frac{m}{x}}=0，则f(0)=0
+\\
+若k\neq 0，则\lim_{x\rightarrow 0}f(x)=\lim_{x\rightarrow 0}\ln \left [ \left ( 1+kx\right )^{\frac{k}{x}}\right ]^{\frac{m}{k}}
+\\
+=\lim_{x\rightarrow 0}\ln e^{\frac{m}{k}}=\frac{m}{k}，则f(0)=\frac{m}{k}
+\\
+.解:k=1，因为此时\lim_{x\rightarrow 0^{-}}f(x)=\lim_{x\rightarrow 0^{-}}\frac{1}{x}\sin x=1
+\\
+\lim_{x\rightarrow 0^{+}}f(x)=\lim_{x\rightarrow 0^{+}}\left ( x\sin \frac{1}{x}+1\right )=1
+\\
+有\lim_{x\rightarrow 0^{+}}f(x)=\lim_{x\rightarrow 0^{-}}f(x)=\lim_{x\rightarrow 0}f(x)=1=f(0)，所以f(x)在x=0点连续
+\\
+由初等函数的连续性可得，f(x)在(-\infty ,0)和(0,+\infty)上连续
+\\
+所以此时f(x)在其定义域R上连续
+\\
+.证明：设f(x)=x^{5}-3x-1，则f(1)=-3，f(2)=25
+\\
+f(1)\cdot f(2)< 0
+\\
+由初等函数的连续性可得f(x)在(1,2)上连续，所以f(x)在(1,2)上至少存在一个零点，故得证。
+\\
+.证明：令y=f(x),则f(1)=-5,f(2)=8
+\\
+f(1)\cdot f(2)< 0
+\\
+由初等函数的连续性可得f(x)在(1,2)上连续，故得证。
+\\
+.证明：f(0)=-1,f(2)=e^{2}-2
+\\
+f(1)\cdot f(2)=2-e^{2}<0
+\\
+由初等函数的连续性可得f(x)在(0,2)上连续，故得证。
+\\
+.(1)解：原式=\frac{\lim_{x\rightarrow 0}\ln \left ( 1+x^{2}\right )}{\lim_{x\rightarrow 0}\sin \left ( 1+x^{2}\right )}=\frac{\ln 1}{\sin 1}=0
+\\
+(2)解：原式=\lim_{x\rightarrow 0}\sqrt{\frac{\lg 100}{a^{x}}}
+\\
+若a\neq 0,则\lim_{x\rightarrow 0}\left ( \frac{\lg 100+x}{a^{x}+\arcsin x}\right )^{\frac{1}{2}}=\sqrt{\frac{\lg 100}{1}}=\sqrt{2}
+\\
+若a=0,则\lim_{x\rightarrow 0}\left ( \frac{\lg 100+x}{a^{x}+\arcsin x}\right )^{\frac{1}{2}}={\infty }
+\\
+</math>
+----
+<math>
+\\
+.(3)
+\\
+解：y^{'}=9\left ( 3x+5\right )^{2}\left ( 5x+4\right )^{5}+25\left ( 3x+5\right )^{3}\left ( 5x+4\right )^{4}
+\\
+=\left [ 9\left ( 5x+4\right )+25\left ( 3x+5\right )\right ]\left ( 3x+5\right )^{2}\left ( 5x+4\right )^{4}
+\\
+=3\left ( 40x+37\right )\left ( 3x+5\right )^{2}\left ( 5x+4\right )^{4}
+\\
+(6)解：y^{'}=\frac{1}{2\sqrt{x^{2}-a^{2}}}\cdot  2x
+\\
+=\frac{x}{\sqrt{x^{2}-a^{2}}}
+\\
+(7)解：y^{'}=\frac{\sqrt{1-x^{2}}-x\cdot \frac{1}{2\sqrt{1-x^{2}}}\cdot \left ( -1\right )\cdot 2x}{1-x^{2}}
+\\
+=\frac{\sqrt{1-x^{2}}+\frac{x^{2}}{\sqrt{1-x^{2}}}}{1-x^{2}}
+\\
+=\frac{1}{\sqrt{1-x^{2}}\cdot 1-x^{2}}
+\\
+=\frac{\sqrt{1-x^{2}}}{\left (  1-x^{2}\right )^{2}}
+\\
+(8)解：y^{'}=\frac{1}{\left ( 1+x^{2}\right )\cdot \ln a}\cdot 2x
+\\
+\frac{2x}{\left ( 1+x^{2}\right )\cdot \ln a}
+\\
+(17)解：y^{'}=\frac{1}{2\cos ^{2}\frac{x}{2}}-\frac{1}{2}
+\\
+(18)解：y^{'}=\frac{1}{\tan \frac{x}{2}}\cdot \frac{1}{\cos ^{2}\frac{x}{2}}\cdot \frac{1}{2}
+\\
+\frac{1}{2\sin \frac{x}{2}\cos \frac{x}{2}}
+\\
+=\frac{1}{\sin x}
+\\
+(19)解：y^{'}=2x\cdot \sin \frac{1}{x}+x^{2}\cdot +\cos \frac{1}{x}\cdot \left ( -1\right )\cdot \frac{1}{x^{2}}
+\\
+=2x\cdot \sin \frac{1}{x}-\cos \frac{1}{x}
+\\
+(20)解：y^{'}=\frac{1}{\ln x}\cdot \frac{1}{x}
+\\
+=\frac{1}{x\ln x}
+\\
+.(5)
+\\
+解：y^{'}=2\arcsin \frac{x}{2}\cdot \frac{1}{\sqrt{1-\frac{1}{x}}}\cdot \frac{1}{2}
+\\
+=\arcsin \frac{x}{2}\cdot \frac{\sqrt{1-\frac{1}{x}}}{1-\frac{1}{x}}
+\\
+=\arcsin \frac{x}{2}\cdot \frac{x\sqrt{\frac{x-1}{x}}}{x-1}
+\\
+=\frac{\arcsin \frac{x}{2}\cdot \sqrt{x^{2}-x}}{x-1}
+\\
+(6)解：y^{'}=\sqrt{1-x^{2}}+x\cdot \frac{1}{2\sqrt{1-x^{2}}}\cdot \left ( -1\right )\cdot 2x+\frac{1}{\sqrt{1-x^{2}}}
+\\
+=\sqrt{1-x^{2}}+\frac{1}{\sqrt{1-x^{2}}}-\frac{x^{2}}{\sqrt{1-x^{2}}}
+\\
+=2\sqrt{1-x^{2}}
+</math>