Ai giải giúp mình với

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Ai giải giúp mình với
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Hải Đăng 5 months 2021-05-17T11:49:10+00:00 1 Answers 3 views 0

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    2021-05-17T11:50:30+00:00

    Đáp án:

    $\lim\limits_{n\to \infty}\left[\cos\left(\dfrac1n\right) + a\sin\left(\dfrac1n\right)\right]^n= e^a$

    Giải thích các bước giải:

    $\quad \lim\limits_{n\to \infty}\left[\cos\left(\dfrac1n\right) + a\sin\left(\dfrac1n\right)\right]^n$

    $= \lim\limits_{n\to \infty}e^{\displaystyle{\ln\left[\cos\left(\dfrac1n\right) + a\sin\left(\dfrac1n\right)\right]^n}}$

    $= e^{\displaystyle{\lim\limits_{n\to \infty}\ln\left[\cos\left(\dfrac1n\right) + a\sin\left(\dfrac1n\right)\right]^n}}$

    Xét $\lim\limits_{n\to \infty}\ln\left[\cos\left(\dfrac1n\right) + a\sin\left(\dfrac1n\right)\right]^n$

    $= \lim\limits_{n\to \infty}n\ln\left[\cos\left(\dfrac1n\right) + a\sin\left(\dfrac1n\right)\right]$

    $= \lim\limits_{n\to \infty}\dfrac{\ln\left[\cos\left(\dfrac1n\right) + a\sin\left(\dfrac1n\right)\right]}{\dfrac1n}$

    $= \lim\limits_{n\to \infty}\dfrac{\dfrac{\sin\left(\dfrac1n\right) – a\cos\left(\dfrac1n\right)}{n^2\left[\cos\left(\dfrac1n\right) + a\sin\left(\dfrac1n\right)\right]}}{-\dfrac{1}{n^2}}$

    $= \lim\limits_{n\to \infty}\dfrac{a\cos\left(\dfrac1n\right)-\sin\left(\dfrac1n\right)}{\cos\left(\dfrac1n\right) + a\sin\left(\dfrac1n\right)}$

    $= \dfrac{a\cos0 – \sin0}{a\sin0 + \cos0}$

    $= a$

    Do đó:

    $e^{\displaystyle{\lim\limits_{n\to \infty}\ln\left[\cos\left(\dfrac1n\right) + a\sin\left(\dfrac1n\right)\right]^n}} = e^a$

    Hay $\lim\limits_{n\to \infty}\left[\cos\left(\dfrac1n\right) + a\sin\left(\dfrac1n\right)\right]^n= e^a$

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