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How to evaluate the following limit? $$\lim_{x\rightarrow0+}(\cos x +|x|+x)^{\cot x}$$

I tried substitution for $\cot x$ but got nowhere. The result should be $e^2$

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  • $\begingroup$ it is equivalent to $e^{\lim_{x \to 0}\cot(x)(\ln(\cos(x)+|x|+x)}$ and doing case work for $x\geq 0$ or $x<0$ $\endgroup$ – Dr. Sonnhard Graubner Nov 6 '14 at 23:22
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Since $x>0$, \begin{eqnarray*} (\cos x+|x|+x)^{\cot x} & = & (\cos x+2x)^{\frac{\cos x}{\sin x}}\\ & = & \exp\left(\frac{\cos x}{\sin x}\log(\cos x+2x)\right)\\ & = & \exp\left(\frac{x}{\sin x}\cos x\frac{\log(\cos x+2x)}{x}\right). \end{eqnarray*} By L'Hospital's method, $$ \lim_{x\to0+}\frac{\log(\cos x+2x)}{x}=\lim_{x\to0+}\frac{\frac{-\sin x+2}{\cos x+2x}}{1}=2. $$ It is known that $$\lim_{x\to0}\frac{x}{\sin x}=1,\quad \lim_{x\to 0}\cos x=1.$$So as $x\to 0$, we have \begin{eqnarray*} (\cos x+|x|+x)^{\cot x} & = & \exp\left(\frac{x}{\sin x}\cos x\frac{\log(\cos x+2x)}{x}\right)\to \exp(2). \end{eqnarray*}

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