Selesaikan limit x mendekati integral berikut!
[tex] \displaystyle \rm \large {\boxed{ \color{white}{\sf \huge \lim_{x \to (\int_{0}^{1}1 + 2x + 3 {x}^{2}-3)} (\frac{x}{ \sqrt{4 + {2}^{3} x} - \sqrt{4 + {2}^{2}x } }) }}}[/tex]
#stress&bosen
[tex] \displaystyle \rm \large {\boxed{ \color{white}{\sf \huge \lim_{x \to (\int_{0}^{1}1 + 2x + 3 {x}^{2}-3)} (\frac{x}{ \sqrt{4 + {2}^{3} x} - \sqrt{4 + {2}^{2}x } }) }}}[/tex]
#stress&bosen
Verified answer
Jawaban:
[tex]\displaystyle \rm \large {\boxed{ \color{white}{\sf \huge \lim_{x \to (\int_{0}^{1}1 + 2x + 3 {x}^{2}-3) } (\frac{x}{ \sqrt{4 + {2}^{3} x} - \sqrt{4 + {2}^{2}x } }) }}}[/tex]
[tex]x \to \: \int_{0}^{1} (1 + 2x + 3 {x}^{2} -3) \\ [/tex]
[tex] \int( 1+ 2x + 3 {x}^{2} - 3) \: dx \\ \int( 3{x}^{2} + 2x - 2) \: dx \\ = \frac{3 {x}^{2 + 1} }{2 + 1} + \frac{2 {x}^{1 +1 } }{1 + 1} - 2x + C \\ = \frac{3 {x}^{3} }{3} + \frac{2 {x}^{2} }{2} - 2x + C \\ = {x}^{3} + {x}^{2} - 2x + C[/tex]
batas atas' 1 and batas bawah' 0
[tex] = ( {x}^{3} + {x}^{2} - 2x) - ( {x}^{3} + {x}^{2} - 2x) \\ = ( {1}^{3} + {1}^{2} - 2(1)) - ( {0}^{3} + {0}^{2} - 2(0)) \\ (1 + 1 - 2) - (0) = 0[/tex]
Jadi*
[tex]x \to \: \int_{0}^{1} (1 + 2x + 3 {x}^{2} -3) \\ x \to \: 0[/tex]
[tex] lim_{x \to \: 0} \: \frac{x}{ \sqrt{4 + {2}^{3} x} - \sqrt{4 + {2}^{2}x } } \\
[/tex]
[tex] lim_{x \to \: 0} \: \frac{x}{ \sqrt{4 + 8 x} - \sqrt{4 + 4x } } \\
[/tex]
[tex] \frac{x}{ \sqrt{8x + 4} - \sqrt{4x + 4} } . \frac{ \sqrt{8x + 4} + \sqrt{4x + 4} }{ \sqrt{8x + 4} + \sqrt{4x + 4} } \\ [/tex]
[tex] \frac{x( \sqrt{8x + 4} + \sqrt{4x + 4} )}{(8x + 4) - (4x + 4)} = \frac{x( \sqrt{8x + 4} + \sqrt{4x + 4)} }{4x} \\ [/tex]
[tex] \frac{x( \sqrt{8x + 4} + \sqrt{4x + 4} )}{x(4)} = \frac{ \sqrt{8x + 4} + \sqrt{4x + 4} }{4} \\ [/tex]
substitusi nilai x → 0*
[tex] \frac{ \sqrt{8x + 4} + \sqrt{4x + 4} }{4} = \frac{ \sqrt{8(0) + 4} + \sqrt{4(0) + 4} }{4} \\ \frac{ \sqrt{4} + \sqrt{4} }{4} = \frac{4}{4} = 1[/tex]