3

Could anyone explain to me how to calculate the limit $\lim_{x\to0} \sqrt[x]{1+\sin x}$? I am pretty sure that the answer is $e$, but how do I arrive at that? I tried to calculate one-sided limits playing with the Squeeze Theorem: in a neighbourhood $(0,\frac{\pi}{2})$ of $0$ one has $\sin x \leq x$, hence $\sqrt[x]{1+\sin x}\leq \sqrt[x]{1+x}$, and the right-hand side tends do $e$ when $x\to 0$. But how to evaluate the expression from the bottom?

Yiyuan Lee
  • 14,435
a student
  • 63
  • 2

2 Answers2

3

Since $x^y$ is continuous near $x=e$ and $y=1$: $$ \begin{align} \lim_{x\to0}\left(1+\sin(x)\right)^{1/x} &=\lim_{x\to0}\left((1+\sin(x))^{1/\sin(x)}\right)^{\lim\limits_{x\to0}\sin(x)/x}\\ &=e^1 \end{align} $$

Another approach: take logs $$ \begin{align} \log\left(\lim_{x\to0}\left(1+\sin(x)\right)^{1/x}\right) &=\lim_{x\to0}\frac{\sin(x)}{x}\frac1{\sin(x)}\log(1+\sin(x))\\ &=\lim_{x\to0}\frac{\sin(x)}{x}\lim_{x\to0}\frac1{\sin(x)}\log(1+\sin(x))\\ &=\lim_{x\to0}\frac{\sin(x)}{x}\log\left(\lim_{x\to0}(1+\sin(x))^{1/\sin(x)}\right)\\[6pt] &=1\log(e)\\[12pt] &=1 \end{align} $$

robjohn
  • 345,667
  • Thank you for the answer! Do you now of a way of proving this without using continuity? – a student Mar 19 '14 at 08:42
  • 2
    If you don't know the functions are continuous, then you can't even prove that $\lim\limits_{x\to0}(1+\sin(x))^{1/\sin(x)}=e$. Continuity is at the core of limits. I am not sure what you want when you ask to show this without using continuity. The only thing more basic is to give a delta-epsilon proof. – robjohn Mar 19 '14 at 08:46
  • @astudent: I have added a second approach. It still uses continuity, but maybe in a form you find more acceptable. – robjohn Mar 19 '14 at 09:01
0

$$\lim_{x\to0}\left(1+rx\right)^{1/x}=e^r$$

Let $\frac{\sin(x)}{x}=r$ and evaluate

$$\lim_{r\to1}\left(\lim_{x\to0}\left(1+rx\right)^{1/x}\right)=\lim_{r\to1}e^r=e$$

Guy
  • 8,857
  • 1
  • 28
  • 57
  • Your reasoning is incorrect and/or not justified enough. You can't just take the limit as $x \to 0$ and ignore the $r$ just because you didn't write it as $r(x)$... Here, let me try it: I want to compute $\lim_{x \to 0} \frac{x}{x}$. Let $r = x$, then $\lim{r \to 0} \left( \lim_{x \to 0} \frac{x}{r} \right) = \lim_{r \to 0} 0 = 0$. Obviously false... – Najib Idrissi Mar 19 '14 at 08:32
  • @nik $x/x$ isn't continuous, $x^y$ is. there's a difference. – Guy Mar 19 '14 at 08:36
  • Thank you for the answer. But do you perhaps know of a way of proving this without using continuity? – a student Mar 19 '14 at 08:43
  • @astudent why shouldn't we use? – Guy Mar 19 '14 at 08:45