Task number: 2984

Determine \( \displaystyle \lim_{x\to 0}\frac{e^x-1}{x} \) with the assumption that \( \displaystyle e=\lim_{n\to \infty}\left(1+\frac1n\right)^n \)

  • Hint

    Express \(e\) as a limit as \(x\) approaches 0.

    It is better to work with the limit using the inverse function – the natural logarithm.

  • Resolution

    First we should observe that

    \( \displaystyle \lim_{x\to 0}\frac{e^x-1}{x} = \left(\lim_{x\to 0}\frac{\ln(x+1)}{x}\right)^{-1} \).

    Letting \(x=\frac 1n\) we have

    \( \displaystyle e= \lim_{n\to \infty}\left(1+\frac1n\right)^n= \lim_{x\to \infty}\left(1+\frac1x\right)^x= \lim_{x\to 0^+}\left(1+x\right)^{\frac1x} \)

    Be aware that in this step we replaced the sequence \(\left(1+\frac1n\right)^n\) with the function \(\left(1+\frac1x\right)^x\), so we must verify e.g. the continuity and monotonicity of this function on \((1,\infty)\) to be sure that the limit of the resulting function exists.

    But we need a two-sided limit, so

    \( \displaystyle \lim_{x\to 0^-}\left(1+x\right)^{\frac1x}= \lim_{x\to 0^+}\left(1-x\right)^{-\frac1x}= \left(\lim_{x\to 0^+}\left(1-x\right)^{\frac1x}\right)^{-1}= \left(e^{-1}\right)^{-1}=e. \)

    Here we used a similar argument as above, except we started with the limit of the sequence \( \lim_{n\to \infty}\left(1-\frac1n\right)^n=e^{-1} \), which we can of course deduce from \( \lim_{n\to \infty}\left(1+\frac1n\right)^n=e \).

    From here we obtain

    \( \displaystyle \lim_{x\to 0}\frac{\ln(x+1)}{x}= \lim_{x\to 0}\ln\left(\exp\left(\frac{\ln(x+1)}{x}\right)\right)= \ln\left(\lim_{x\to 0}\left(1+x\right)^{\frac1x}\right)=\ln e =1 \)

    Now we must only compute

    \( \displaystyle \lim_{x\to 0}\frac{e^x-1}{x} = 1^{-1}=1 \).

    and also \( \displaystyle \lim_{x\to 0^-}\frac{e^x-1}{x}= \lim_{x\to 0^-}\frac{\left(\left(1-x\right)^{-\frac1x}\right)^x-1}{x}= \lim_{x\to 0^-}\frac{\frac1{1-x}-1}{x}= \lim_{x\to 0^-}\frac{1-(1-x)}{(1-x)x}=1 \)

  • Result

    The limit equals 1.

Difficulty level: Hard task
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