10+6=16 карандашей в двух коробках.
На сколько больше карандашей в первой коробке, чем во второй?
10-6=4 - на столько карандашей больше в первой коробке.
Неопределённость 0/0 раскрывается по правилу Лопиталя или умножением числителя и знаменателя на сопряжённое выражение.
1)
![ \lim_{x \to \inft0} \frac{x}{\sqrt{x+1}-1} = \lim_{x \to \inft0} \frac{x*(\sqrt{x+1}+1)}{(\sqrt{x+1}-1)*(\sqrt{x+1}+1)} = \\ \\ = \lim_{x \to \inft0} \frac{x*(\sqrt{x+1}+1)}{x+1-1} =\lim_{x \to \inft0} (\sqrt{x+1}+1)= \sqrt{0+1} +1= 2](https://tex.z-dn.net/?f=%0A+%5Clim_%7Bx+%5Cto+%5Cinft0%7D++%5Cfrac%7Bx%7D%7B%5Csqrt%7Bx%2B1%7D-1%7D+%3D+%5Clim_%7Bx+%5Cto+%5Cinft0%7D++%0A%5Cfrac%7Bx%2A%28%5Csqrt%7Bx%2B1%7D%2B1%29%7D%7B%28%5Csqrt%7Bx%2B1%7D-1%29%2A%28%5Csqrt%7Bx%2B1%7D%2B1%29%7D+%3D+%5C%5C++%5C%5C+%3D+%0A%5Clim_%7Bx+%5Cto+%5Cinft0%7D++%5Cfrac%7Bx%2A%28%5Csqrt%7Bx%2B1%7D%2B1%29%7D%7Bx%2B1-1%7D+%3D%5Clim_%7Bx+%5Cto+%5Cinft0%7D%0A+%28%5Csqrt%7Bx%2B1%7D%2B1%29%3D++%5Csqrt%7B0%2B1%7D+%2B1%3D+2)
или (по Лопиталю)
![\lim_{x \to \inft0} \frac{x}{\sqrt{x+1}-1} = \lim_{x \to \inft0} \frac{x'}{(\sqrt{x+1}-1)'} =\lim_{x \to \inft0} \frac{1}{ \frac{1}{2} (x+1)^{- \frac{1}{2} }} = \\ \\ = \lim_{x \to \inft0} 2 (x+1)^{ \frac{1}{2}} =\lim_{x \to \inft0} 2 \sqrt{x+1} =2 \sqrt{0+1} =2](https://tex.z-dn.net/?f=%5Clim_%7Bx%0A+%5Cto+%5Cinft0%7D+%5Cfrac%7Bx%7D%7B%5Csqrt%7Bx%2B1%7D-1%7D+%3D+%5Clim_%7Bx+%5Cto+%5Cinft0%7D+%0A%5Cfrac%7Bx%27%7D%7B%28%5Csqrt%7Bx%2B1%7D-1%29%27%7D+%3D%5Clim_%7Bx+%5Cto+%5Cinft0%7D+%5Cfrac%7B1%7D%7B+%5Cfrac%7B1%7D%7B2%7D+%0A%28x%2B1%29%5E%7B-+%5Cfrac%7B1%7D%7B2%7D+%7D%7D+%3D+%5C%5C++%5C%5C+%3D+%5Clim_%7Bx+%5Cto+%5Cinft0%7D+2++%28x%2B1%29%5E%7B+%0A%5Cfrac%7B1%7D%7B2%7D%7D+%3D%5Clim_%7Bx+%5Cto+%5Cinft0%7D+2+%5Csqrt%7Bx%2B1%7D+%3D2+%5Csqrt%7B0%2B1%7D+%3D2)
2)
![ \lim_{x \to \inft0} \frac{\sqrt{x+4}-2}{x} = \lim_{x \to \inft0} \frac{(\sqrt{x+4}-2)*(\sqrt{x+4}+2)}{x*(\sqrt{x+4}+2)} = \\ \\ =\lim_{x \to \inft0} \frac{x+4-4}{x*(\sqrt{x+4}+2)} =\lim_{x \to \inft0} \frac{1}{\sqrt{x+4}+2} = \frac{1}{ \sqrt{0+4} +2} = \frac{1}{4}](https://tex.z-dn.net/?f=%0A+%5Clim_%7Bx+%5Cto+%5Cinft0%7D++%5Cfrac%7B%5Csqrt%7Bx%2B4%7D-2%7D%7Bx%7D+%3D+%5Clim_%7Bx+%5Cto+%5Cinft0%7D+%0A%5Cfrac%7B%28%5Csqrt%7Bx%2B4%7D-2%29%2A%28%5Csqrt%7Bx%2B4%7D%2B2%29%7D%7Bx%2A%28%5Csqrt%7Bx%2B4%7D%2B2%29%7D+%3D+%5C%5C++%5C%5C+%3D%5Clim_%7Bx%0A+%5Cto+%5Cinft0%7D+%5Cfrac%7Bx%2B4-4%7D%7Bx%2A%28%5Csqrt%7Bx%2B4%7D%2B2%29%7D+%3D%5Clim_%7Bx+%5Cto+%5Cinft0%7D+%0A%5Cfrac%7B1%7D%7B%5Csqrt%7Bx%2B4%7D%2B2%7D+%3D+%5Cfrac%7B1%7D%7B+%5Csqrt%7B0%2B4%7D+%2B2%7D+%3D+%5Cfrac%7B1%7D%7B4%7D+)
или (по Лопиталю)
![\lim_{x \to \inft0} \frac{\sqrt{x+4}-2}{x} =\lim_{x \to \inft0} \frac{(\sqrt{x+4}-2)'}{x'} = \lim_{x \to \inft0} \frac{1}{2} (x+4)^{- \frac{1}{2} } = \\ \\ =\lim_{x \to \inft0} \frac{1}{2 \sqrt{x+4} }= \frac{1}{2* \sqrt{0+4} } = \frac{1}{4}](https://tex.z-dn.net/?f=%5Clim_%7Bx%0A+%5Cto+%5Cinft0%7D+%5Cfrac%7B%5Csqrt%7Bx%2B4%7D-2%7D%7Bx%7D+%3D%5Clim_%7Bx+%5Cto+%5Cinft0%7D+%0A%5Cfrac%7B%28%5Csqrt%7Bx%2B4%7D-2%29%27%7D%7Bx%27%7D+%3D+%5Clim_%7Bx+%5Cto+%5Cinft0%7D+%5Cfrac%7B1%7D%7B2%7D+%28x%2B4%29%5E%7B-+%0A%5Cfrac%7B1%7D%7B2%7D+%7D+%3D+%5C%5C++%5C%5C+%3D%5Clim_%7Bx+%5Cto+%5Cinft0%7D+%5Cfrac%7B1%7D%7B2+%5Csqrt%7Bx%2B4%7D+%7D%3D+%0A%5Cfrac%7B1%7D%7B2%2A+%5Csqrt%7B0%2B4%7D+%7D+%3D+%5Cfrac%7B1%7D%7B4%7D++)
3)
![ \lim_{x \to \inft0} \frac{x}{\sqrt{x+1}-1} =\lim_{x \to \inft0} \frac{x*(\sqrt{x+1}+1)}{(\sqrt{x+1}-1)*(\sqrt{x+1}+1)} = \\ \\ =\lim_{x \to \inft0} \frac{x*(\sqrt{x+1}+1)}{x+1-1} = \lim_{x \to \inft0} (\sqrt{x+1}+1) = (\sqrt{0+1}+1) = 2](https://tex.z-dn.net/?f=+%0A%5Clim_%7Bx+%5Cto+%5Cinft0%7D++%5Cfrac%7Bx%7D%7B%5Csqrt%7Bx%2B1%7D-1%7D+%3D%5Clim_%7Bx+%5Cto+%5Cinft0%7D++%0A%5Cfrac%7Bx%2A%28%5Csqrt%7Bx%2B1%7D%2B1%29%7D%7B%28%5Csqrt%7Bx%2B1%7D-1%29%2A%28%5Csqrt%7Bx%2B1%7D%2B1%29%7D+%3D+%5C%5C++%5C%5C+%3D%5Clim_%7Bx%0A+%5Cto+%5Cinft0%7D+%5Cfrac%7Bx%2A%28%5Csqrt%7Bx%2B1%7D%2B1%29%7D%7Bx%2B1-1%7D+%3D+%5Clim_%7Bx+%5Cto+%5Cinft0%7D+%0A%28%5Csqrt%7Bx%2B1%7D%2B1%29+%3D+%28%5Csqrt%7B0%2B1%7D%2B1%29+%3D+2)
или (по Лопиталю)
![\lim_{x \to \inft0} \frac{x}{\sqrt{x+1}-1} =\lim_{x \to \inft0} \frac{x'}{(\sqrt{x+1}-1)'} = \lim_{x \to \inft0} \frac{1}{ \frac{1}{2} (x+1)^{- \frac{1}{2}}}= \\ \\ =\lim_{x \to \inft0} 2 (x+1)^{ \frac{1}{2}} =\lim_{x \to \inft0} 2 \sqrt{x+1} = 2 \sqrt{0+1} =2](https://tex.z-dn.net/?f=%5Clim_%7Bx%0A+%5Cto+%5Cinft0%7D+%5Cfrac%7Bx%7D%7B%5Csqrt%7Bx%2B1%7D-1%7D+%3D%5Clim_%7Bx+%5Cto+%5Cinft0%7D+%0A%5Cfrac%7Bx%27%7D%7B%28%5Csqrt%7Bx%2B1%7D-1%29%27%7D+%3D+%5Clim_%7Bx+%5Cto+%5Cinft0%7D+%5Cfrac%7B1%7D%7B+%5Cfrac%7B1%7D%7B2%7D+%0A%28x%2B1%29%5E%7B-+%5Cfrac%7B1%7D%7B2%7D%7D%7D%3D++%5C%5C++%5C%5C+%3D%5Clim_%7Bx+%5Cto+%5Cinft0%7D+2++%28x%2B1%29%5E%7B+%0A%5Cfrac%7B1%7D%7B2%7D%7D+%3D%5Clim_%7Bx+%5Cto+%5Cinft0%7D+2+%5Csqrt%7Bx%2B1%7D+%3D+2+%5Csqrt%7B0%2B1%7D+%3D2)
Ответ:
1)6*2=12(см) длина
2)Р=(6+12)*2=36(см)
3)S=6*12=72(см2) (в квадрате)
3^x=a
a+10/a-11<0
a²-11a+10<0
a1+a2=11 U a1*a2=10
a1=1 U a2=10
1<a<10
1<3^x<10
0<x<log(3)10
x∈(0;log(3)10)