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Good evening to everyone, the limit is $\lim\limits _{x\to \infty }\left(\left(1-\left|x\right|\right)e^{\frac{1}{2x+2}}\right)$ and I don't know how to compute it I tried the limit substitution, the $ e^{\ln x} $ technique and I tried to transform this case of $0\cdot \infty$ in a case of $\frac{\infty }{\infty }$ or $ \frac{0}{0} $ but I don't know how. Thanks for any possible answers.

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$\lim_\limits{x\to \infty }\left(\left(1-\left|x\right|\right)e^{\frac{1}{2x+2}}\right)=\lim_\limits{x\to \infty }\left(1-\left|x\right|\right)\lim_\limits{x\to \infty }e^{\frac{1}{2x+2}}\\ \lim_\limits{x\to \infty }e^{\frac{1}{2x+2}} = 1\\ \lim_\limits{x\to \infty }\left(1-\left|x\right|\right) = -\infty$

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  • $\begingroup$ Yeah, thanks. I didn't observe that. xD $\endgroup$
    – T4yl0r
    Jun 15 '16 at 19:57

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