Toán Lớp 12: Tìm tiệm cận ngang của hs:
y=(x+1)².(2x-1)³/3x⁵-5x+7
y=(x+1)³.(x+2)⁴/(x+3)⁵.(x²+1)
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Toán Lớp 12: Tìm tiệm cận ngang của hs:
y=(x+1)².(2x-1)³/3x⁵-5x+7
y=(x+1)³.(x+2)⁴/(x+3)⁵.(x²+1)
Comments ( 1 )
1)\quad y = \dfrac{(x+1)^2(2x-1)^3}{3x^5 – 5x + 7}\\
\text{Ta có:}\\
\quad \lim\limits_{x\to \pm \infty}y\\
= \lim\limits_{x\to \pm \infty}\dfrac{(x+1)^2(2x-1)^3}{3x^5 – 5x + 7}\\
= \lim\limits_{x\to \pm \infty}\dfrac{\left(1 + \dfrac{1}{x}\right)^2\left(2 – \dfrac{1}{x}\right)^3}{3 – \dfrac{5}{x^4} + \dfrac{7}{x^5}}\\
= \dfrac{(1+0)^2(2 – 0)^3}{3 – 0 + 0}\\
= \dfrac83\\
\Rightarrow y = \dfrac83\ \text{là TCN của đồ thị hàm số}\\
2)\quad y = \dfrac{(x+1)^3(x+2)^4}{(x+3)^5(x^2 + 1)}\\
\text{Ta có:}\\
\quad \lim\limits_{x\to \pm \infty}y\\
= \lim\limits_{x\to \pm \infty}\dfrac{(x+1)^3(x+2)^4}{(x+3)^5(x^2 + 1)}\\
= \lim\limits_{x\to \pm \infty}\dfrac{\left(1 + \dfrac{1}{x}\right)^3\left(1 + \dfrac{2}{x}\right)^3}{\left(1 + \dfrac{3}{x}\right)^5\left(1 + \dfrac{1}{x^2}\right)}\\
= \dfrac{(1 + 0)^3(1 + 0)^4}{(1 + 0)^5(1+0)}\\
= 1\\
\Rightarrow y = 1\ \text{là TCN của đồ thị hàm số}\\
\end{array}\)