помогите с математикой . Created by rayahvan746. matematika-ru

помогите с математикой ...

rayahvan746 @rayahvan746

August 2022 1 17 Report

помогите с математикой

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sangers1959

Пошаговое объяснение:

$a)\ \lim_{x \to 4} \frac{\sqrt{x+5}-3 }{x-4}=\lim_{x \to 4} \frac{(\sqrt{x+5}-3 )'}{(x-4)'}= \lim_{x \to 4} \frac{\frac{1}{2\sqrt{x+5} } }{1} = \lim_{x \to 4} \frac{1}{2*\sqrt{x+5} }=\\
 =\frac{1}{2*\sqrt{4+5} }=\frac{1}{2\sqrt{9} }=\frac{1}{2*3}=\frac{1}{6}.\\
$

$b)\ \lim_{x \to 2} \frac{x-2}{\sqrt{x+2}-2 } = \lim_{x \to 2} \frac{(x-2)'}{(\sqrt{x+2}-2)' } = \lim_{x \to 2}\frac{1}{\frac{1}{2\sqrt{x+2} } } = \lim_{x \to 2}( 2*\sqrt{x+2})=\\
 =2*\sqrt{2+2}=2*\sqrt{4} =2*2=4.$

$c)\ \lim_{x \to 1} \frac{2\sqrt{x+3}-4 }{x-1}= \lim_{x \to 1} \frac{(2\sqrt{x+3}-4)' }{(x-1)'}= \lim_{x \to 1} \frac{\frac{2}{2\sqrt{x+3} } }{1} = \lim_{n \to 1} \frac{1}{\sqrt{x+3} }=\\
 = \frac{1}{\sqrt{1+3} } =\frac{1}{\sqrt{4} } =\frac{1}{2} .$

$d)\ \lim_{x \to3 } \frac{x-3}{\sqrt{3x+7}-4 }= \lim_{x \to 3 } \frac{(x-3)'}{(\sqrt{3x+7}-4 )'}= \lim_{x \to 3} \frac{1}{\frac{1}{2\sqrt{3x+7} } }= \lim_{x \to 3} 2\sqrt{3x+7} =\\
 =2*\sqrt{3*3+7}=2*\sqrt{9+7}=2*\sqrt{16} =2*4=8.$

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rayahvan746 большое спасибо)❤️

sangers1959 Удачи.

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