Tìm giới hạn \(C=\lim \limits_{x \rightarrow+\infty}\left[\sqrt[n]{\left(x+a_{1}\right)\left(x+a_{2}\right) \ldots\left(x+a_{n}\right)}-x\right]:\)
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Lời giải:
Báo saiĐặt \(y=\sqrt[n]{\left(x-a_{1}\right)\left(x-a_{2}\right) \ldots\left(x-a_{n}\right)}\)
\(\begin{array}{l} \Rightarrow y^{n}-x^{n}=(y-x)\left(y^{n-1}+y^{n-1} x+\ldots+x^{n-1}\right) \Rightarrow y-x=\frac{y^{n}-x^{n}}{y^{n-1}+y^{n-1} x+\ldots+x^{n-1}} \\ \Rightarrow \lim \limits_{x \rightarrow+\infty}(y-x)=\lim\limits _{x \rightarrow+\infty} \dfrac{y^{n}-x^{n}}{y^{n-1}+y^{n-2} x+\ldots+x^{n-1}} \\ \Rightarrow C=\lim\limits _{x \rightarrow+\infty} \dfrac{\frac{y^{n}-x^{n}}{x^{n-1}}}{\frac{y^{n-1}+y^{n-1} x+\ldots+x^{n-1}}{x^{n-1}}} . \end{array}\)
\(\begin{array}{l} \text { Mà } \lim\limits _{x \rightarrow+\infty} \frac{y^{n}-x^{n}}{x^{n-1}}=\lim \limits _{x \rightarrow+\infty}\left(a_{1}+a_{2}+\ldots+a_{n}+\frac{b_{2}}{x}+\frac{b_{3}}{x^{2}}+\ldots+\frac{b_{n}}{x^{n-1}}\right) \\ =a_{1}+a_{2}+\ldots+a_{n} \\ \lim\limits _{x \rightarrow+\infty} \frac{y^{k} x^{n-1-k}}{x^{n-1}}=1 \forall k=0, \ldots, n-1 \Rightarrow \lim \limits _{x \rightarrow+\infty} \frac{y^{n-1}+y^{n-2} x+\ldots+x^{n-1}}{x^{n-1}}=n . \\ \text { Vậy } C=\frac{a_{1}+a_{2}+\ldots+a_{n}}{n} \end{array}\)
