Multiple Choice Question (MCQ)

\lim\limits_{x\to 0} \dfrac{(1+x)^{\tan x}-1}{x^2}=


  1. 1

  2. ×

    2

  3. ×

    -1

  4. ×

    -2

Collected in the board: Limit

Steven Zheng posted 2 weeks ago

Answer

  1. \lim\limits_{x\to 0} \dfrac{(1+x)^{\tan x}-1}{x^2}

    =\lim\limits_{x\to 0} \dfrac{e^{\ln(1+x)^{\tan x}}-1}{x^2}

    =\lim\limits_{x\to 0} \dfrac{\tan x\ln(1+x)}{x^2}

    =\lim\limits_{x\to 0} \dfrac{\sin x}{x}\dfrac{\ln(1+x) }{x}

    =1

    So A is the right choice.

    In the last step, we used two basic limits

    \lim\limits_{x\to 0} \dfrac{\sin x}{x} = 1

    and

    \lim\limits_{x\to 0} \dfrac{\ln(1+x) }{x} = 1

    For the second one,

    \lim\limits_{x\to 0} \dfrac{\ln(1+x) }{x}

    = \lim\limits_{x\to 0} \ln(1+x)^{\frac{1}{x}}

    = \lim\limits_{x\to 0} \ln e

    =1

Steven Zheng posted 2 weeks ago

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