$\lim_{x \to \infty} \frac{3^x+2^{x-1}}{4^x}$ 27/10/2021 Bởi Jasmine $\lim_{x \to \infty} \frac{3^x+2^{x-1}}{4^x}$
$\displaystyle\lim_{x \to \infty} \dfrac{3^x+2^{x-1}}{4^x}\\ =\displaystyle\lim_{x \to \infty} \dfrac{3^x+\dfrac{1}{2}.2^{x}}{4^x}\\ =\displaystyle\lim_{x \to \infty} \dfrac{\left(\dfrac{3}{4}\right)^x+\dfrac{1}{2}.\left(\dfrac{2}{4}\right)^x}{1}\\ =0$ Bình luận
$\displaystyle\lim_{x \to \infty} \dfrac{3^x+2^{x-1}}{4^x}\\ =\displaystyle\lim_{x \to \infty} \dfrac{3^x+\dfrac{1}{2}.2^{x}}{4^x}\\ =\displaystyle\lim_{x \to \infty} \dfrac{\left(\dfrac{3}{4}\right)^x+\dfrac{1}{2}.\left(\dfrac{2}{4}\right)^x}{1}\\ =0$
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