cho hàm số f(x)= (4x+1)^3 (2x+1)^4 / (3+2x)^7 .tính lim x-> âm vô cùng f(x)

$\begin{array}{l}
\mathop {\lim }\limits_{x \to  - \infty } f\left( x \right) = \mathop {\lim }\limits_{x \to  - \infty } \dfrac{{{{\left( {4x + 1} \right)}^3}{{\left( {2x + 1} \right)}^4}}}{{{{\left( {3 + 2x} \right)}^7}}}\\
 = \mathop {\lim }\limits_{x \to  - \infty } \dfrac{{{x^3}{{\left( {4 + \dfrac{1}{x}} \right)}^3}{x^4}.{{\left( {2 + \dfrac{1}{x}} \right)}^4}}}{{{x^7}{{\left( {2 + \dfrac{3}{x}} \right)}^7}}}\\
 = \mathop {\lim }\limits_{x \to  - \infty } \dfrac{{{{\left( {4 + \dfrac{1}{x}} \right)}^3}{{\left( {2 + \dfrac{1}{x}} \right)}^4}}}{{{{\left( {2 + \dfrac{3}{x}} \right)}^7}}} = \dfrac{{{4^3}{{.2}^4}}}{{{2^7}}} = \dfrac{{{2^6}{{.2}^4}}}{{{2^7}}} = {2^3} = 8
\end{array}$

 Đáp án:

$2^{3}$

Giải thích các bước giải:

$\lim_{x \to -\infty} f(x)$ = $\frac{(4x+1)^{3}.(2x+1)^{4}}{(3+2x)^{7}}$ = $\frac{(4+\frac{1}{x})^{3}.(2+\frac{1}{x})^{4}}{(\frac{3}{x}+2)^{7}}$ = $\frac{(4+0)^{3}.(2+0)^{4}}{(0+2)^{7}}$ = $2^{3}$