1в) используем второй замечательный предел
![= \lim_{x \to \infty} (1+ \frac{1}{x^2} ) ^{x^2+1}= \lim_{x \to \infty} (1+ \frac{1}{x^2} ) ^{x^2}(1+ \frac{1}{x^2} )=e*1=e](https://tex.z-dn.net/?f=%3D+%5Clim_%7Bx+%5Cto+%5Cinfty%7D+%281%2B+%5Cfrac%7B1%7D%7Bx%5E2%7D+%29+%5E%7Bx%5E2%2B1%7D%3D+%5Clim_%7Bx+%5Cto+%5Cinfty%7D+%281%2B+%5Cfrac%7B1%7D%7Bx%5E2%7D+%29+%5E%7Bx%5E2%7D%281%2B+%5Cfrac%7B1%7D%7Bx%5E2%7D+%29%3De%2A1%3De)
2б)
![y = e^{ \sqrt[5]{2x} *ln(arctg(x/3))}\\y'=(arctg( \frac{x}{3} ))^{\sqrt[5]{2x}}*( \frac{\sqrt[5]{2}}{5 \sqrt[5]{x^4}} ln(arctg(x/3)+ \frac{\sqrt[5]{2x}}{3arctg(x/3)(1+x^2/9)} )](https://tex.z-dn.net/?f=y+%3D+e%5E%7B+%5Csqrt%5B5%5D%7B2x%7D+%2Aln%28arctg%28x%2F3%29%29%7D%5C%5Cy%27%3D%28arctg%28+%5Cfrac%7Bx%7D%7B3%7D+%29%29%5E%7B%5Csqrt%5B5%5D%7B2x%7D%7D%2A%28++%5Cfrac%7B%5Csqrt%5B5%5D%7B2%7D%7D%7B5+%5Csqrt%5B5%5D%7Bx%5E4%7D%7D++ln%28arctg%28x%2F3%29%2B++%5Cfrac%7B%5Csqrt%5B5%5D%7B2x%7D%7D%7B3arctg%28x%2F3%29%281%2Bx%5E2%2F9%29%7D+%29)
4a)
![= \lim_{x \to +0} \frac{ln(1+x)-ln(1+x^2)}{ln(1+x^2)ln(1+x)} = \\=\lim_{x \to +0} ( \frac{1}{1+x}- \frac{2x}{1+x^2} ):(\frac{2xln(1+x)}{1+x^2}+ \frac{ln(1+x^2)}{1+x} )= \frac{1}{+0}= +\infty](https://tex.z-dn.net/?f=%3D+%5Clim_%7Bx+%5Cto+%2B0%7D++%5Cfrac%7Bln%281%2Bx%29-ln%281%2Bx%5E2%29%7D%7Bln%281%2Bx%5E2%29ln%281%2Bx%29%7D++%3D+%5C%5C%3D%5Clim_%7Bx+%5Cto+%2B0%7D+%28+%5Cfrac%7B1%7D%7B1%2Bx%7D-+%5Cfrac%7B2x%7D%7B1%2Bx%5E2%7D++%29%3A%28%5Cfrac%7B2xln%281%2Bx%29%7D%7B1%2Bx%5E2%7D%2B+%5Cfrac%7Bln%281%2Bx%5E2%29%7D%7B1%2Bx%7D+%29%3D+%5Cfrac%7B1%7D%7B%2B0%7D%3D+%2B%5Cinfty)
4б)
![= \lim_{x \to 0} e^{1/x^2 * ln( \frac{arctgx}{x} )} =e^{ \lim_{x \to 0} \frac{ln(arctgx/x)}{x^2} }=\\=e^{ \lim_{x \to 0} \frac{1}{arctgx(1+x^2)*2x} }=e^{\infty}=\infty](https://tex.z-dn.net/?f=%3D+%5Clim_%7Bx+%5Cto+0%7D+e%5E%7B1%2Fx%5E2+%2A+ln%28+%5Cfrac%7Barctgx%7D%7Bx%7D+%29%7D+%3De%5E%7B+%5Clim_%7Bx+%5Cto+0%7D++%5Cfrac%7Bln%28arctgx%2Fx%29%7D%7Bx%5E2%7D++%7D%3D%5C%5C%3De%5E%7B+%5Clim_%7Bx+%5Cto+0%7D++%5Cfrac%7B1%7D%7Barctgx%281%2Bx%5E2%29%2A2x%7D++%7D%3De%5E%7B%5Cinfty%7D%3D%5Cinfty)
Две пятых минус одна третья будет ...и одна вторая 1
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2
три дели на одну вторую
ответ 6 целых
1%-5
2%-x
x=2*5/1=10
1%-5
6%-x
x=6*5/1=30
1%-5
100%-x
x=5*100/1=500
3(65+А)-РАССТОЯНИЕ МЕЖДУ ПОЕЗДАМИ
а=10 3*(65+10)=225 км
а=25 3*(65+25)=270км
а=40 3*(65+40)= 315 км