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I am unable to solve this easy limit without using L'Hospital, can you help me and maybe explain how can I solve it?

$$\lim_{x \rightarrow -\infty} e^x \log|x|$$

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    $\begingroup$ I suggest writing it as the equivalent: $\lim_{x\to\infty}e^{-x}\log x$ $\endgroup$
    – GPerez
    Commented Dec 13, 2014 at 16:50

5 Answers 5

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$$\lim\limits_{x \to -\infty}e^x\log{|x|}=\lim\limits_{x \to -\infty}\frac{\log{|x|}}{e^{-x}}=\lim\limits_{x \to \infty}\frac{\log{x}}{e^x}=0$$

ps : you can use \to in $\LaTeX$ for $\to$.

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  • $\begingroup$ Thank you a lot! I didn't know about the trick of changing sign to infinity. Is it something that I can do every time that I got a $\lim\limits_{x \to -\infty}$ ? $\endgroup$
    – linkxvi
    Commented Dec 13, 2014 at 17:17
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    $\begingroup$ Well, it's just substituting the variable. To be more clear, you can specify it : $\lim\limits_{x \to - \infty}x = \lim\limits_{x' \to \infty}-x' = \lim\limits_{x'' \to 0^-}\frac{1}{x''}$ (these are the most common but you can substitute with other functions) . Yes it will work basically with any expression. Other example : $\lim\limits_{x \to \pi} sin^2(x)= \lim\limits_{x' \to 1}x'^2$ $\endgroup$
    – servabat
    Commented Dec 13, 2014 at 17:24
  • $\begingroup$ Hmm that's actually $x \to 0$ for sin sorry.. $\endgroup$
    – servabat
    Commented Dec 13, 2014 at 17:32
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$f(n)=e^{-n}\log n$
If $n\geq2$ then $n+1<n^2$, so $\log(n+1)<2\log n$.
So $0<f(n+1)<\frac 2ef(n)$.

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$\lim_{x\rightarrow -\infty}e^x\log|x|=\lim_{x\rightarrow \infty}e^{-x}\log(x)\leq \lim_{x\rightarrow \infty}e^{-x}x=0$ ($e^x$ increases faster than every polynomial)

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Hint. Assume $\{x_n\} = -e^n$.

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It's difficult to answer a question like this without knowing what we are allowed to use. It's tempting to write

$$\lim_{x\to-\infty}e^x\log|x|=\lim_{x\to\infty}\left({x\over e^x}\cdot{\log x\over x} \right)$$

and invoke the "known" limits

$$\lim_{x\to\infty}\left(x\over e^x\right)=\lim_{n\to\infty}\left(\log x\over x\right)=0$$

If these aren't considered known, then we can reduce the first to the second by letting $x=\log u$ and prove the second from the definition of the natural logarithm,

$${\log x\over x}={1\over x}\int_1^x {dt\over t}={1\over x}\int_1^\sqrt x{dt\over t}+{1\over x}\int_\sqrt x^x{dt\over t}\le{\sqrt x\over x}+{1\over\sqrt x}\to0$$

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