$\displaystyle \lim_{x \to +\infty} \frac {x^4}{e^x}$ $\displaystyle \lim_{x \to +\infty} \frac {\ln x}{x}$ Can anyone show me how these limit reaches to 0?
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Originally Posted by chengbin $\displaystyle \lim_{x \to +\infty} \frac {x^4}{e^x}$ $\displaystyle \lim_{x \to +\infty} \frac {\ln x}{x}$ Can anyone show me how these limit reaches to 0? Use l'Hospital's rule 4 times for the first limit and once for the second.
Originally Posted by chengbin $\displaystyle \lim_{x \to +\infty} \frac {x^4}{e^x}$ $\displaystyle \lim_{x \to +\infty} \frac {\ln x}{x}$ Can anyone show me how these limit reaches to 0? $\displaystyle \lim_{x \to +\infty} \frac {x^4}{e^x} = 0 = \lim_{x \to +\infty} \frac {\ln x}{x}$, because you know that $\displaystyle x^4 < e^x \ \& \ ln x < x$ for big x Proof by using L'Hospital
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