1. limit x mendekati ∞ : x²-4x+1 per 2x²+7x-5 2. limit x mendekati ∞ : 8x³-3x²-4x+9 per 4xpangkat5+2.... Question from @ditajulia

Takamori37 Nomor 1.
$$\begin{align}\lim_{x\to\infty}\frac{x^2-4x+1}{2x^2+7x-5}&=\lim_{x\to\infty}\frac{x^2(1-\frac{4}{x}+\frac{1}{x^2})}{x^2(2+\frac{7}{x}-\frac{5}{x^2})} \\ &=\lim_{x\to\infty}\frac{1-\frac{4}{x}+\frac{1}{x^2}}{2+\frac{7}{x}-\frac{5}{x^2}} \\ &=\frac{1-0+0}{2+0-0} \\ &=\frac{1}{2}\end{align}$

Nomor 2.
$$\begin{align}\lim_{x\to\infty}\frac{8x^3-3x^2-4x+9}{4x^5+2x^3-2x-3}&=\lim_{x\to\infty}\frac{x^5(\frac{8}{x^2}-\frac{3}{x^3}-\frac{4}{x^4}+\frac{9}{x^5})}{x^5(4+\frac{2}{x^2}-\frac{2}{x^4}-\frac{3}{x^5})} \\ &=\lim_{x\to\infty}\frac{\frac{8}{x^2}-\frac{3}{x^3}-\frac{4}{x^4}+\frac{9}{x^5}}{4+\frac{2}{x^2}-\frac{2}{x^4}-\frac{3}{x^5}} \\ &=\frac{0-0-0+0}{4+0-0-0} \\ &=\frac{0}{4} \\ &=0\end{align}$

Nomor 3.
$$\begin{align}\lim_{x\to\infty}\frac{\sqrt{4x^2-2x+5}}{x-5}&=\lim_{x\to\infty}\frac{x(\frac{1}{x}\sqrt{4x^2-2x+5})}{x(1-\frac{5}{x})} \\ &=\lim_{x\to\infty}\frac{\sqrt{4-\frac{2}{x}+\frac{5}{x^2}}}{1-\frac{5}{x}} \\ &=\frac{\sqrt{4-0+0}}{1-0} \\ &=\frac{\sqrt{4}}{1}=\frac{2}{1} \\ &=2\end{align}$

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ditajulia tolong dong bantuin jawab pertanyaan aku yang lain

ditajulia masih tentang limit

echavezjane $1. \frac{x^2-4x+1}{2x^2+7x-5} = \frac{\frac{1}{x^2}}{\frac{1}{x^2}} \cdot \frac{x^2-4x+1}{2x^2+7x-5} = \frac{1 - \frac{4}{x} + \frac{1}{x^2}}{2+\frac{7}{x}-\frac{5}{x^2}}\\ \lim_{x\to\infty} \frac{x^2-4x+1}{2x^2+7x-5} = \lim_{x\to\infty} \frac{1 - \frac{4}{x} + \frac{1}{x^2}}{2+\frac{7}{x}-\frac{5}{x^2}} = \frac{1}{2}.$

$2. \frac{8x^3-3x^2-4x+9}{4x^5 +2x^3 -2x-3} = \frac{\frac{1}{x^5}}{\frac{1}{x^5}} \cdot \frac{8x^3-3x^2-4x+9}{4x^5 +2x^3 -2x-3} = \frac{\frac{8}{x^2}-\frac{3}{x^3}-\frac{4}{x^4}+\frac{9}{x^5}}{4+\frac{2}{x^2}-\frac{2}{x^4}-\frac{3}{x^5}} \\ \lim_{x\to\infty}\frac{8x^3-3x^2-4x+9}{4x^5 +2x^3 -2x-3} = \lim_{x\to\infty}\frac{\frac{8}{x^2}-\frac{3}{x^3}-\frac{4}{x^4}+\frac{9}{x^5}}{4+\frac{2}{x^2}-\frac{2}{x^4}-\frac{3}{x^5}} = \frac{0}{4} = 0.$

$3. \frac{ \sqrt{4x^2 - 2x+5} }{x-5} = \frac{\frac{1}{x}}{\frac{1}{x}}\cdot \frac{ \sqrt{4x^2 - 2x+5} }{x-5} \\ =\frac{ \sqrt{4-\frac{2}{x} + \frac{5}{x^2}} }{1-\frac{5}{x}}$ karena $\frac{1}{x} = \frac{1}{ \sqrt{x^2}}$ .

$\lim_{x\to\infty} \frac{ \sqrt{4x^2 - 2x+5} }{x-5} = \lim_{x\to\infty} \frac{ \sqrt{4-\frac{2}{x}+\frac{5}{x^2}} }{1-\frac{5}{x}} = \frac{ \sqrt{4} }{1} = 2.$

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