Find the value of :$\lim_{x \rightarrow 0^+} \frac{\ln x}{x^n} \rightarrow -\infty$?

Question

How to figure out that $\lim_{x \rightarrow 0^+} \frac{\ln x}{x^n} \rightarrow -\infty$ and not $\infty$, because according to L'Hopitals:

$$\lim_{x \rightarrow 0^+} \frac{\ln x}{x^n} = \lim_{x \rightarrow 0^+} \frac{1/x}{nx^{n-1}}=\lim_{x \rightarrow 0^+} \frac{1}{nx^{n}} \rightarrow \infty$$

You can't directly use LHopital since top initially goes to $-\infty$ while bottom goes to $0^+$ (assuming $n>0.$) — coffeemath, May 01 '16 at 07:37

score 2 · Accepted Answer · edited Apr 13 '17 at 12:19

2

First: you do not have to use L'Hospital's rule when it is not necessary. Second, you should not when the hypotheses do not apply. You can see a discussion on a related case at Why does L'Hopital's rule fail in this case?.

At $0^+$, $\frac{1}{x^n}\to +\infty$ (for $n>0$) and $\log x \to -\infty$, and the form $+\infty \times -\infty$ is not undeterminate, it yields here $-\infty$.

Why does L'Hopital's rule fail in this case?

edited Apr 13 '17 at 12:19

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1

answered May 01 '16 at 07:36

Laurent Duval

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How do I know when it is appropriate to use it? :) – Naz May 01 '16 at 07:38
When the form is underterminate: $0/0$ or $\pm \infty/\pm \infty$. It is a temping one, but unnecessary in many cases – Laurent Duval May 01 '16 at 07:41
I think you answered for inappropriate there haha. Ok, thanks – Naz May 01 '16 at 07:43
I do not understand. It is unappropriate to use it when the form is determinate, and it might be appropriate when it is underterminate ($0/0$ for instance) – Laurent Duval May 01 '16 at 07:46

Find the value of :$\lim_{x \rightarrow 0^+} \frac{\ln x}{x^n} \rightarrow -\infty$?

1 Answers1