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Compute using L'Hopital's rule:

$$\lim_{x\to 0^+} \frac{\ln(x)}{1/\sin(x)}$$

I kept differentiating, but it's getting too long. How can I tackle this kind of problem?

Also, when I encounter a limit of the form $\infty \cdot 0$ and I want to make it eligible for L'Hopital's rule, should I transform it into $$\frac{\infty}{1/0}=\frac{\infty}{\infty}$$ or $$\frac{0}{1/\infty}=\frac{0}{0}$$

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3 Answers 3

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Or without l'Hopital: $$\frac{\ln x}{\frac1{\sin x}} = (x\ln x)\cdot \frac{\sin x}x,$$ where each factor has a well-known limit as $x\to0^+$.

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You chose the poorer way to turn the $0\cdot\infty$ form into one to which l’Hospital’s rule applies.

$$\begin{align*} \lim_{x\to 0^+}\ln x\sin x&=\lim_{x\to 0^+}\frac{\sin x}{1/\ln x}\\\\ &=\lim_{x\to 0^+}\frac{\cos x}{-(\ln x)^{-2}\cdot\frac1x}\\\\ &=-\lim_{x\to 0^+}\frac{(\ln x)^2\cos x}{1/x}\;. \end{align*}$$

At this point you can try try applying l’Hospital’s rule again, but it’s clear that the numerator is going to be fairly messy. A better idea is to notice that $\lim\limits_{x\to 0^+}\cos x= 1$, so that

$$-\lim_{x\to 0^+}\frac{(\ln x)^2\cos x}{1/x}=-\left(\lim_{x\to 0^+}\frac{(\ln x)^2}{1/x}\right)\lim_{x\to 0^+}\cos x=-\lim_{x\to 0^+}\frac{(\ln x)^2}{1/x}\;;$$ this gets rid of the trig function. Now

$$\begin{align*} -\lim_{x\to 0^+}\frac{(\ln x)^2}{1/x}&=-\lim_{x\to 0^+}\frac{\frac2x\ln x}{-1/x^2}\\\\ &=2\lim_{x\to 0^+}\frac{\ln x}{1/x}\\\\ &=2\lim_{x\to 0^+}\frac{1/x}{-1/x^2}\\\\ &=-2\lim_{x\to 0^+}x\\\\ &=0\;. \end{align*}$$

It’s always a good idea to keep your eyes open for factors with known finite, non-zero limits, like the $\cos x$ above: generally speaking, it’s a good idea to simplify as much as possible the expression whose limit you’re taking.

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  • $\begingroup$ so in my question, is there a way of telling which is better before trying it? $\endgroup$
    – Maria
    Oct 27, 2012 at 20:14
  • $\begingroup$ @user1561559: Sometimes it doesn’t matter much. When it does, there’s no foolproof way to tell ahead of time, though experience can often give you a pointer. $\endgroup$ Oct 27, 2012 at 20:17
  • $\begingroup$ @M.Strochyk: You’re right: I don’t know how I lost that minus sign in the exponent. $\endgroup$ Oct 27, 2012 at 21:05
  • $\begingroup$ @M.Strochyk: Fixed now, and the corrected version has the virtue of making another useful point. $\endgroup$ Oct 27, 2012 at 21:21
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When you use LHospital's rule $$\lim_{x \to 0} \dfrac{\ln(x)}{\csc(x)} = \lim_{x \to 0} \dfrac{\left(\dfrac{d\ln(x)}{dx} \right)}{\left(\dfrac{d\csc(x)}{dx} \right)} = \lim_{x \to 0} \dfrac{\left(1/x \right)}{\left(-\cot(x) \csc(x) \right)} = -\lim_{x \to 0} \dfrac{\sin(x) \tan(x)}{x}\\ = - \left(\lim_{x \to 0} \dfrac{\sin(x)}{x} \right) \times \left(\lim_{x \to 0} \tan(x) \right)$$

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  • $\begingroup$ well this gives 0/0? $\endgroup$
    – Maria
    Oct 27, 2012 at 20:04
  • $\begingroup$ Do you know $\lim_{x\to 0} \frac{\sin(x)}{x}$? $\endgroup$ Oct 27, 2012 at 20:04
  • $\begingroup$ @user1561559 What is $\lim_{x \to 0} \sin(x)/x$ and $\lim_{x \to 0} \tan(x)$? $\endgroup$
    – user17762
    Oct 27, 2012 at 20:06
  • $\begingroup$ ok yea i get it.. thanks $\endgroup$
    – Maria
    Oct 27, 2012 at 20:11

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