MTK~Limit _______________ [tex]\Large\underline{\sf{MATEMATIKA}}[/tex] [tex]\large\sf{lim_{x\to\infty}\ \left(2\sqrt{x}+1\right)-\sqrt{4x-3\sqrt{x}+2}=...}[/tex] [tex]\small\boxed{\tt{Note_{1}=Jawablah\ dengan\ usaha\ sendiri}}[/tex] [tex]\small\boxed{\tt{Note_{2}=Dilarang\ nyalin\ jawaban\ maupun\ copas\ dari\

Berikut ini adalah pertanyaan dari Sinogen pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

MTK~Limit_______________

\Large\underline{\sf{MATEMATIKA}}



\large\sf{lim_{x\to\infty}\ \left(2\sqrt{x}+1\right)-\sqrt{4x-3\sqrt{x}+2}=...}




\small\boxed{\tt{Note_{1}=Jawablah\ dengan\ usaha\ sendiri}}
\small\boxed{\tt{Note_{2}=Dilarang\ nyalin\ jawaban\ maupun\ copas\ dari\ web}}
\small\boxed{\tt{Note_{3}=Jaga \ Kesehatan \ yah}}



Terimakasih ^^
MTK~Limit
_______________
[tex]\Large\underline{\sf{MATEMATIKA}}[/tex]
[tex]\large\sf{lim_{x\to\infty}\ \left(2\sqrt{x}+1\right)-\sqrt{4x-3\sqrt{x}+2}=...}[/tex]
[tex]\small\boxed{\tt{Note_{1}=Jawablah\ dengan\ usaha\ sendiri}}[/tex]
[tex]\small\boxed{\tt{Note_{2}=Dilarang\ nyalin\ jawaban\ maupun\ copas\ dari\ web}}[/tex]
[tex]\small\boxed{\tt{Note_{3}=Jaga \ Kesehatan \ yah}}[/tex]
Terimakasih ^^

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Jawaban:

\begin{aligned}\lim_{x\to\infty}\:\left(\left(2\sqrt{x}+1\right)-\sqrt{4x-3\sqrt{x}+2}\right)=\bf\frac{7}{4}\end{aligned}

Pembahasan

Limit

\begin{aligned}&\lim_{x\to\infty}\:\left(\left(2\sqrt{x}+1\right)-\sqrt{4x-3\sqrt{x}+2}\right)\\&{=\ }\lim_{x\to\infty}\:\left(2\sqrt{x}+1-\sqrt{4x-3\sqrt{x}+2}\right)\\&{=\ }\lim_{x\to\infty}\:1+\lim_{x\to\infty}\:\left(2\sqrt{x}-\sqrt{4x-3\sqrt{x}+2}\right)\\&{=\ }1+\lim_{x\to\infty}\:\left[\left(2\sqrt{x}-\sqrt{4x-3\sqrt{x}+2}\right)\cdot\frac{2\sqrt{x}+\sqrt{4x-3\sqrt{x}+2}}{2\sqrt{x}+\sqrt{4x-3\sqrt{x}+2}}\right]\end{aligned}

\begin{aligned}&{=\ }1+\lim_{x\to\infty}\:\left(\frac{4x-4x+3\sqrt{x}-2}{2\sqrt{x}+\sqrt{4x-3\sqrt{x}+2}}\right)\\&{=\ }1+\lim_{x\to\infty}\:\left(\frac{3\sqrt{x}-2}{2\sqrt{x}+\sqrt{4x-3\sqrt{x}+2}}\right)\\&{=\ }1+\lim_{x\to\infty}\:\left(\frac{3\sqrt{x}}{2\sqrt{x}+\sqrt{4x-3\sqrt{x}+2}}\right)-\lim_{x\to\infty}\:\left(\frac{2}{2\sqrt{x}+\sqrt{4x-3\sqrt{x}+2}}\right)\\&{=\ }1+\lim_{x\to\infty}\:\left(\frac{3\sqrt{x}}{2\sqrt{x}+\sqrt{4x-3\sqrt{x}+2}}\right)-0\end{aligned}

\begin{aligned}&{=\ }1+\lim_{x\to\infty}\:\left(\frac{3\sqrt{x}}{2\sqrt{x}+\sqrt{4x-3\sqrt{x}+2}}\right)\\&{=\ }1+3\cdot\lim_{x\to\infty}\:\left(\frac{\sqrt{x}}{2\sqrt{x}+\sqrt{4x-3\sqrt{x}+2}}\right)\\&{=\ }1+3\cdot\lim_{x\to\infty}\:\left(\frac{\sqrt{x}}{2\sqrt{x}+\sqrt{4x-3\sqrt{x}+2}}\cdot\frac{\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{x}}}\right)\end{aligned}

\begin{aligned}&{=\ }1+3\cdot\lim_{x\to\infty}\:\left(\frac{\dfrac{\cancel{\sqrt{x}}}{\cancel{\sqrt{x}}}}{ \dfrac{2\cancel{\sqrt{x}}}{\cancel{\sqrt{x}}} +\dfrac{\sqrt{4x-3\sqrt{x}+2}}{\sqrt{x}}}\right)\\&{=\ }1+3\cdot\lim_{x\to\infty}\:\left(\frac{1}{2 +\sqrt{\dfrac{4x-3\sqrt{x}+2}{x}}}\right)\\&{=\ }1+3\cdot\lim_{x\to\infty}\:\left(\frac{1}{2 +\sqrt{4-\frac{3}{\sqrt{x}}+\frac{2}{x}}}\right)\end{aligned}

\begin{aligned}&{=\ }1+3\cdot\frac{\lim\limits_{x\to\infty}1}{\lim\limits_{x\to\infty}2+\lim\limits_{x\to\infty}\:\sqrt{4-\frac{3}{\sqrt{x}}+\frac{2}{x}}}\\&{=\ }1+3\cdot\frac{\lim\limits_{x\to\infty}1}{\lim\limits_{x\to\infty}2+\sqrt{\lim\limits_{x\to\infty}\left(4-\frac{3}{\sqrt{x}}+\frac{2}{x}\right)}}\\&{=\ }1+3\cdot\frac{1}{2+\sqrt{4-0+0}}\\&{=\ }1+3\cdot\frac{1}{2+2}\ =\ 1+\frac{3}{4}\\&{=\ }\bf\frac{7}{4}\end{aligned}

Kesimpulan

\begin{aligned}\therefore\ \boxed{\ \lim_{x\to\infty}\:\left(\left(2\sqrt{x}+1\right)-\sqrt{4x-3\sqrt{x}+2}\right)=\bf\frac{7}{4}\ }\end{aligned}

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Last Update: Wed, 08 Jun 22