Find $A=\lim_{x \rightarrow 0 }(\frac{(1+x)^{1/x} - e + \frac{ex}{2}}{x^2})$

Question

Question:

Find the value $$A=\lim_{x \rightarrow 0 }(\frac{(1+x)^{1/x} - e + \frac{ex}{2}}{x^2})$$

I found the value $$A=\lim_{x \rightarrow 0 }(\frac{(1+x)^{1/x} - e + \frac{ex}{2}}{x^2})=\lim_{x \rightarrow 0 }( \frac{e- e + ex/2}{x^2}) =0 $$

Is this correct ?

No. That first step is not right but even if it was, how would that last limit be $0$...? — StackTD, Dec 05 '18 at 08:50
https://math.stackexchange.com/questions/447429/how-to-solve-this-limit-lim-limits-x-to0-frac1x1-x-ex — lab bhattacharjee, Dec 05 '18 at 08:51
You mean the last step? After applying it once, the numerator becomes a constant but the denominator is still $2x$... — StackTD, Dec 05 '18 at 08:52
I assume OP mistook the limit as being to infinity rather than 0 — PrincessEev, Dec 05 '18 at 08:53

score 3 · Answer 1 · answered Dec 05 '18 at 09:02

3

Assuming that you know Taylor series $$a=(1+x)^{\frac 1x}\implies \log(a)={\frac 1x}\log(1+x)={\frac 1x}\left(x-\frac{x^2}{2}+\frac{x^3}{3}+O\left(x^4\right) \right)$$ that is to say $$\log(a)=1-\frac{x}{2}+\frac{x^2}{3}+O\left(x^3\right)$$ Continue with Taylor using $$a=e^{\log(a)}=e-\frac{e }{2}x+\frac{11 e }{24}x^2+O\left(x^3\right)$$

answered Dec 05 '18 at 09:02

Claude Leibovici

260,315

@ [email protected] im not getting ur last sentence $a = e^{loga}$ can u elaboarte that how e - e/2x +......my question is that how $e$ come ? – jasmine Dec 05 '18 at 09:10

Find $A=\lim_{x \rightarrow 0 }(\frac{(1+x)^{1/x} - e + \frac{ex}{2}}{x^2})$

1 Answers1