Question

tính \(\lim\limits_{x\rightarrow+\infty}\) \(\dfrac{\left(x+1\right)\left(x^2+1\right)\left(x^3+1\right)...\left(x^{11}+1\right)}{[\left(11x\right)^{11}+1]^6}\)

Answer 1

Lời giải:

Ta chia cả tử cả mẫu cho \(x^{66}\):

\(\lim_{x\to +\infty}\frac{(x+1)(x^2+1)(x^3+1)...(x^{11}+1)}{[11x)^{11}+1]^6}=\lim_{x\to +\infty}\frac{\frac{(x+1)(x^2+1)(x^3+1)....(x^{11}+1)}{x^{66}}}{\frac{[(11x)^{11}+1]^6}{x^{66}}}\)

\(=\lim_{x\to +\infty}\frac{\left ( \frac{x+1}{x} \right )\left ( \frac{x^2+1}{x^2} \right )...\left ( \frac{x^{11}+1}{x^{11}} \right )}{\left [ \frac{(11x)^{11}+1}{x^{11}} \right ]^6}=\lim_{x\to +\infty}\frac{\left ( 1+\frac{1}{x} \right )\left ( 1+\frac{1}{x^2} \right )....\left ( 1+\frac{1}{x^{11}} \right )}{\left ( 11^{11}+\frac{1}{x^{11}} \right )^6}\)

\(=\frac{1.1...1}{(11^{11})^6}=\frac{1}{11^{66}}\)