DENGAN CARA: 1. Nilai dari Lim x -> tak terhingga ((2x - 1) - akar 4x^2 - 6x + 5) 2. Nilai dari Lim .... Question from @kriskrisma

October 2018 1 117 Report

DENGAN CARA:

1. Nilai dari Lim x -> tak terhingga ((2x - 1) - akar 4x^2 - 6x + 5)

2. Nilai dari Lim x -> tak terhingga (akar 4x^2 - 8x + 3 - 2x - 4)

3. Nilai dari Lim x -> 3 x^2 - 9 / akar 10 + 2x - (x + 1)

Takamori37 Nomor 1 dan 2 dapat menggunakan rumus cepatnya.
Dengan:
1.)
$$\begin{align}&\lim_{x\to\infty}2x-1-\sqrt{4x^2-6x+5}\\&=\lim_{x\to\infty}\sqrt{(2x-1)^2}-\sqrt{4x^2-6x+5} \\ &=\lim_{x\to\infty}\sqrt{\boxed{4}x^2\boxed{-4}x+1}-\sqrt{4x^2\boxed{-6}x+5} \\ &=\frac{\boxed{-4}-\boxed{-6}}{2\sqrt{\boxed{4}}} \\ &=\frac{-4+6}{2\times2}=\frac{2}{4} \\ &=\frac{1}{2}\end{align}$

2.)
$$\begin{align}&\lim_{x\to\infty}\sqrt{4x^2-8x+3}-(2x+4)\\&=\lim_{x\to\infty}\sqrt{4x^2-8x+3}-\sqrt{(2x+4)^2} \\ &=\lim_{x\to\infty}\sqrt{\boxed{4}x^2\boxed{-8}x+3}-\sqrt{4x^2+\boxed{16}x+16} \\ &=\frac{\boxed{-8}-\boxed{16}}{2\sqrt{\boxed{4}}} \\ &=\frac{-8-16}{2\times2}=\frac{-24}{4} \\ &=-6\end{align}$

3.)
$$\begin{align}&\lim_{x\to3}\frac{x^2-9}{\sqrt{10+2x}-(x+1)}\\&=\lim_{x\to3}\frac{x^2-9}{\sqrt{10+2x}-(x+1)}\times\frac{\sqrt{10+2x}+(x+1)}{\sqrt{10+2x}+(x+1)} \\ &=\lim_{x\to3}\frac{(x^2-9)(\sqrt{10+2x}+(x+1))}{10+2x-(x+1)^2} \\ &=\lim_{x\to3}\frac{(x^2-9)(\sqrt{10+2x}+(x+1))}{10+2x-(x^2+2x+1)} \\ &=\lim_{x\to3}\frac{(x^2-9)(\sqrt{10+2x}+(x+1))}{x^2-9} \\ &=\lim_{x\to3}\sqrt{10+2x}+x+1 \\ &=\sqrt{10+2(3)}+3+1 \\ &=4+4\\&=8\end{align}$

6 votes Thanks 27

kriskrisma Kalo bisa tolong jawab soal aku yg lain ya, yg akar 5-4x+3x^2 dst. Thank you ^^