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  1. By finding the limit as $x\to0$ (using Taylor series) $\dfrac{e^x -1}{x}$, I got $e^x = 1 + x + \dfrac{x^2}{2} + ... $ so $e^x -1 = x + \dfrac{x^2}{2} + O(x^2)$. But the derivative of $\dfrac{1}{x}$ gives $-\dfrac{1}{x^2}$ by inserting(while calculating taylor series) $0$ when I try to compute Taylor series of $\dfrac{1}{x}$ it gives me $\dfrac{1}{0}$. can someone check where I made a mistake?
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  • $\begingroup$ What limit? $x\to 0$? $\endgroup$
    – user223391
    Jan 21, 2017 at 15:04
  • $\begingroup$ @ZacharySelk yes x --> 0. $\endgroup$
    – Khan Saab
    Jan 21, 2017 at 15:05
  • $\begingroup$ That is relevant information. :) $\endgroup$
    – user223391
    Jan 21, 2017 at 15:05
  • $\begingroup$ sorry i will edit it again :-) $\endgroup$
    – Khan Saab
    Jan 21, 2017 at 15:06
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    $\begingroup$ $\frac{e^x-1}{x} = 1 + x/2 + O(x^2)$ so the limit as $x$ goes to $0$ is $1$ $\endgroup$
    – Zubzub
    Jan 21, 2017 at 15:07

1 Answer 1

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If you want to evaluate $\lim_\limits{x\to 0}\dfrac{e^x -1}{x}$ using Taylor series, you just need to know $e^x -1 = x + O(x)$. Then you have $\lim_\limits{x\to 0}\dfrac{x}{x}$ which is $1$.

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