Ответы и объяснения

2013-11-03T08:08:13+00:00
A) Можно двумя способами. Второй и все остальные пределы по правилу Лопиталя.     1)   \lim_{x \to 0}  \frac{\sqrt{ x^{2} +x}-x}{x+ x^{2} }=\lim_{x \to 0}\frac{(\sqrt{ x^{2}+x}-x)(\sqrt{ x^{2} +x}+x)}{x(1+ x)(\sqrt{ x^{2}+x}+x) }=\lim_{x \to 0}\frac{x^{2}+x- x^{2}}{x(1+ x)(\sqrt{x^{2}+x}+x) }=
=\lim_{x \to 0}\frac{x}{x(1+ x)(\sqrt{x^{2}+x}+x) }=\lim_{x \to 0}\frac{1}{(1+ x)(\sqrt{x^{2}+x}+x) }= \infty
2)   \lim_{x \to 0} \frac{\sqrt{ x^{2} +x}-x}{x+ x^{2} }=\lim_{x \to 0}\frac{(\sqrt{ x^{2}+x}-x)'}{(x+ x^{2})'  }=\lim_{x \to 0}\frac{ \frac{2x+1}{2\sqrt{ x^{2} +x}} - 1}{2x+1}=
=\lim_{x \to 0}\frac{ 2x+1-2\sqrt{ x^{2} +x}}{2\sqrt{ x^{2} +x}(2x+1)}=  \{ \frac{1}{0} \}= \infty   
b)   \lim_{x \to0} \frac{ln(1+2x)}{tgx}=\lim_{x \to0}\frac{ln'(1+2x)}{tg'x}=\lim_{x \to0}\frac{\frac{2}{1+2x}}{ \frac{1}{cos^2x}}= =\lim_{x \to0}\frac{2cos^2x}{1+2x}}=\frac{2cos^20}{1+2*0}}=2
c)  \lim_{x \to 0} \frac{2^x-3^x}{3^x-4^x}= \lim_{x \to 0} \frac{(2^x-3^x)'}{(3^x-4^x)'}=\lim_{x \to 0} \frac{2^x*ln2-3^x*ln3}{3^x*ln3-4^x*ln4}=\frac{ln2-ln3}{ln3-ln4};
d)   \lim_{x \to  \frac{ \pi }{2}+0 }5^{ \frac{1}{1-cosx}}=5^{ \frac{1}{1-cos \frac{ \pi }{2}}}=5^1=5
e)   \lim_{x \to 0} \frac{ x^{2}+sin5x}{sin(tgx^3)}=\lim_{x \to 0} \frac{( x^{2} +sin5x)'}{sin'(tgx^3)}=\lim_{x \to 0} \frac{2x +5cos5x}{cos(tgx^3)* \frac{1}{cos^2x^3}*3 x^{2}}= =\frac{2*0 +5cos5*0}{cos(tg0)* \frac{1}{cos^20}*3 *0}=\{\frac{5}{1* 1*0}\}=\{\frac{5}{0}\}=\infty
f)   \lim_{x \to-1} \frac{tg(1+x)}{x^{2}-x-2}=\lim_{x \to-1} \frac{tg'(1+x)}{(x^{2}-x-2)'}=\lim_{x \to-1} \frac{ \frac{1}{cos^2(1+x)} }{2x-1}=  =\frac{ \frac{1}{cos^2(1-1)} }{-2-1}= -\frac{1}{3};
g)   \lim_{x \to \infty}\frac{x \sqrt{ x^{2}+1}+\sqrt[3]{x^5-1}}{2 x^{2}-\sqrt[3]{x^7+x^6}}=\lim_{x \to \infty}\frac{x^2( \sqrt{1+ \frac{1}{x^{2}} }+\sqrt[3]{ \frac{1}{x} - \frac{1}{x^6} })}{x^{2}(2-\sqrt[3]{x+1})}=
\lim_{x \to \infty}\frac{ \sqrt{1+ \frac{1}{x^{2}} }+\sqrt[3]{ \frac{1}{x} - \frac{1}{x^6} }}{2-\sqrt[3]{x+1}}=\frac{ \sqrt{1+  0 } + \sqrt[3]{  0  - 0 }}{2-\infty}= \frac{1}{ \infty }=0