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I want to calculate $\lim \limits_{n \to 0^{+}} (\sin x)^{e^{x}-1} $ by using Taylor's Series, and here is what I did so far, and correct me if I'm wrong:

  • $\sin x = x + o(x)$
  • $e^{x}-1= x + o(x)$

then I substituted these two expressions in the initial limit and got something like this:

$\lim \limits_{x \to 0^{+}} (x+o(x))^{x+o(x)}$

but how do I continue from here?

rubik
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Firas Abd El Gani
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3 Answers3

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$$\lim_{x \to 0^{+}} (\sin x)^{e^{x}-1}=\lim_{x\to0^+}e^{\frac{e^x-1}x.x\ln(\sin x)}$$ Now: $$\lim_{x\to0}\frac{e^x-1}x=1\\\lim_{x\to0^+}x\ln(\sin x)=0$$ the last one because: $$\lim_{x\to0^+}x\ln(\sin x)=\lim_{x\to0^+}\frac{\ln(\sin x)}{1/x}=\lim_{x\to0^+}\frac{\cot x}{-1/x^2}=\lim_{x\to0^+}-x.\frac{x}{\tan x}=0$$ as: $$\lim_{x\to0}\frac{\tan x}x=1$$ So limit is $1$.

rubik
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RE60K
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    Ok, first of all thanks for the effort. I really appreciate it. secondly, I wanted to see an example for Taylor series, and Im sure u used l'Hopital's rule here at some point, although I tagged Limits without l'Hopital. – Firas Abd El Gani Mar 02 '15 at 13:02
  • OK @FirasAliAbdelGhani maybe someone else will work out without l'hospital – RE60K Mar 02 '15 at 13:31
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The first order Taylor expansions $\sin(u) \underset{u\to0}= u + o(u)$ and $\ln(1+u) \underset{u\to 0}= u + o(u)$ give for $x \to 0^+$, $$ \log(\sin(x)) = \log(x+o(x)) = \log x + \log(1+o(1)) = \log x + o(1). $$ Therefore, using $e^u \underset{x\to 0} = 1 + x +o(x)$, we obtain \begin{align} (\sin(x))^{e^x - 1} &= \exp[(e^x - 1)\log (\sin(x))]\\ & = \exp[(x+o(x))(\log x +o(1))]\\ & = \exp[x\log x + o(x\log x)]\\ & = \exp[o(1)]\\ & = 1 + o(1) \end{align} since $x\log x = o(1)$.

Siméon
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If $f(x) = (\sin x)^{e^{x} - 1}$ then $f(x)$ is defined in some interval of type $(0, a)$ and hence it makes sense to ask for its limit as $x \to 0^{+}$. If $L$ is the desired limit then \begin{align} \log L &= \log\left(\lim_{x \to 0^{+}}(\sin x)^{e^{x} - 1}\right)\notag\\ &= \lim_{x \to 0^{+}}\log(\sin x)^{e^{x} - 1}\text{ (by continuity of log)}\notag\\ &= \lim_{x \to 0^{+}}(e^{x} - 1)\log \sin x\notag\\ &= \lim_{x \to 0^{+}}\frac{e^{x} - 1}{x}\cdot x\cdot\log \left(x\cdot\frac{\sin x}{x}\right)\notag\\ &= \lim_{x \to 0^{+}}1\cdot x\cdot\left\{\log x + \log\left(\frac{\sin x}{x}\right)\right\}\notag\\ &= \lim_{x \to 0^{+}}x\log x + x\log\left(\frac{\sin x}{x}\right)\notag\\ &= \lim_{x \to 0^{+}}x\log x + 0\cdot\log 1\notag\\ &= \lim_{x \to 0^{+}}x\log x\notag\\ &= 0\notag \end{align} Hence $L = e^{0} = 1$. Here we have used the standard limits $$\lim_{x \to 0}\frac{\sin x}{x} = 1,\,\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1,\, \lim_{x \to 0^{+}}x \log x = 0$$

Paramanand Singh
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