- 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?
Limit calculation with Taylor series
0
$\begingroup$
taylor-expansion
limits-without-lhopital
-
0What limit? $x\to 0$? – 2017-01-21
-
0@ZacharySelk yes x --> 0. – 2017-01-21
-
0That is relevant information. :) – 2017-01-21
-
0sorry i will edit it again :-) – 2017-01-21
-
2$\frac{e^x-1}{x} = 1 + x/2 + O(x^2)$ so the limit as $x$ goes to $0$ is $1$ – 2017-01-21
-
0Do we always need taylor series for both(numerator and denominator) or we can compute taylor series only for $e^x -1$ and then divide both side both(numerator and denominator) by x? – 2017-01-21
-
0As you did by subtracting $1$ on the Taylor expansion of $e^x$ you can also divide the expansion by $x$ ! – 2017-01-21
-
0They Taylor series for $x$ is just $x$ – 2017-01-21
1 Answers
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 $1$.