lim x->1+( sin(1-x)-(e^(x-1))+1)/ lnx

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lim x->1+( sin(1-x)-(e^(x-1))+1)/ lnx

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Thanh Hà 4 years 2021-08-27T10:28:15+00:00 1 Answers 13 views 0

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  1. ngochoa
    0
    2021-08-27T10:29:51+00:00
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    We’re given the one-sided limit,

    \displaystyle\lim_{x\to1^+}\frac{\sin(1-x)-e^{x-1}+1}{\ln(x)}

    Evaluating the limand directly at x = 1 gives the indeterminate from

    (sin(1 – 1) – exp(1 – 1) + 1) / ln(1) = 0/0

    so we can potentially solve the limit by applying L’Hopital’s rule. Doing so gives

    \displaystyle\lim_{x\to1^+}\frac{\sin(1-x)-e^{x-1}+1}{\ln(x)}=\lim_{x\to1^+}\frac{-\cos(1-x)-e^{x-1}}{\frac1x}=\frac{-\cos(0)-e^0}{\frac11}=\boxed{-2}

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