I can't solve this limit without using L'Hospital: $\lim_{x \rightarrow -\infty} e^x \log|x|$ I am unable to solve this easy limit without using L'Hospital, can you help me and maybe explain how can I solve it?
$$\lim_{x \rightarrow -\infty} e^x \log|x|$$
 A: $$\lim\limits_{x \to -\infty}e^x\log{|x|}=\lim\limits_{x \to -\infty}\frac{\log{|x|}}{e^{-x}}=\lim\limits_{x \to \infty}\frac{\log{x}}{e^x}=0$$
ps : you can use \to in $\LaTeX$ for $\to$.
A: $f(n)=e^{-n}\log n$
If $n\geq2$ then $n+1<n^2$, so $\log(n+1)<2\log n$.
So $0<f(n+1)<\frac 2ef(n)$.
A: $\lim_{x\rightarrow -\infty}e^x\log|x|=\lim_{x\rightarrow \infty}e^{-x}\log(x)\leq \lim_{x\rightarrow \infty}e^{-x}x=0$ ($e^x$ increases faster than every polynomial)
A: Hint. Assume $\{x_n\} = -e^n$.
A: 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,
$${\log x\over x}={1\over x}\int_1^x {dt\over t}={1\over x}\int_1^\sqrt x{dt\over t}+{1\over x}\int_\sqrt x^x{dt\over t}\le{\sqrt x\over x}+{1\over\sqrt x}\to0$$
