Limit calculation with Taylor series

Question

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?

Answer 1

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 $1$.