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The limit is: $$ \lim_{\lambda\to 0} \frac{\int_{\lambda}^{a}{\frac{\cos(x)}{x}dx}}{\ln\lambda}. $$

My argument is:

First rewrite the integral: $$ \lim_{\lambda\to 0} \frac{\int_{0}^{a}{\frac{\cos(x)}{x}dx} - \int_{0}^{\lambda}{\frac{\cos(x)}{x}dx}}{\ln\lambda}. $$

Then use l'Hopital's rule. The first term on the top vanishes as it has not $\lambda$ dependence. The second term is found by applying fundamental theorem of calculus. So I get:

$$ \lim_{\lambda\to 0} \frac{-{\frac{\cos(\lambda)}{\lambda}dx}}{\frac{1}{\lambda}} = \lim_{\lambda\to 0}{-\cos(\lambda)} = -1. $$

Are there any problems with my argument?

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Ilya Lapan
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1 Answers1

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There are just two small problems.

You cannot compute the two-sided limit, because $$ \int_{0}^{a}\frac{\cos x}{x}\,dx $$ does not converge. So, assuming $a>0$, the limit can only be computed for $\lambda\to0^+$.

You have $$ \int_{\lambda}^a\frac{\cos x}{x}\,dx= -\int_a^{\lambda}\frac{\cos x}{x}\,dx $$ and the derivative is $$ -\frac{\cos\lambda}{\lambda} $$ without the need to split the integral (by the way, you chose the wrong way to split it, because $\int_0^a\frac{\cos x}{x}\,dx$ does not exist, as said above).

In full detail, for $a>0$, $$ \lim_{\lambda\to0^+} \frac{\displaystyle\int_{\lambda}^a \frac{\cos x}{x}\,dx}{\ln\lambda}= \lim_{\lambda\to0^+} \frac{-\dfrac{\cos \lambda}{\lambda}}{\dfrac{1}{\lambda}}= \lim_{\lambda\to0^+}-\cos\lambda=-1 $$

Note that l’Hôpital can be applied to the forms $\dfrac{\text{whatever}}{\infty}$. However, it's also easy to see that $$ \lim_{\lambda\to0^+}\int_\lambda^a\frac{\cos x}{x}\,dx=\infty $$ because in the interval $[0,\pi/3]$ we have $\cos x\ge\frac{1}{2}$, so $$ \int_\lambda^a\frac{\cos x}{x}\,dx= \int_\lambda^{\pi/3}\frac{\cos x}{x}\,dx+ \int_{\pi/3}^a\frac{\cos x}{x}\,dx\ge \frac{1}{2}\int_\lambda^{\pi/3}\frac{1}{x}\,dx+ \int_{\pi/3}^a\frac{\cos x}{x}\,dx $$ so, by comparison, we get the required limit.

egreg
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  • Yep, I meant approaching zero from the right. I forgot it when I was typing it up. – Ilya Lapan Feb 24 '17 at 21:03
  • Not fully done, yet, but I'm too lazy to type my own answer: an important missing step is to check that the hypotheses of l'Hospital's theorem are verified. In particular, to check that the limit of the numerator is $\pm \infty$. Splitting it as the OP does only produces an indeterminacy of the type $\infty - \infty$. One should, in exchange, apply the limit comparison test for improper integrals (comparing, for instance, with $\int _0 ^a \frac 1 x \ \Bbb d x = \infty$), and only next apply l'Hospital's theorem. – Alex M. Feb 24 '17 at 21:03
  • @AlexM. L’Hôpital’s theorem is also applicable to forms $\text{whatever}/\infty$. However, I added the easy proof. – egreg Feb 24 '17 at 21:12
  • @egreg: Provided that you show the numerator to have a value - finite or infinite. What if the integral were divergent, not infinite (something like $\int _0 ^\infty \cos x \ \Bbb e ^x \ \Bbb d x$, for instance)? This case must be ruled out. – Alex M. Feb 24 '17 at 21:13
  • @AlexM. It doesn't matter, actually, whether the numerator has a limit or not. Not a well known fact, but you can see it in the Wikipedia page about l'Hôpital’s theorem. The example you make is irrelevant: the numerator in this case is well defined for every $\lambda>0$, being the integral of a continuous function (note $a>0$ and $\lambda>0$). – egreg Feb 24 '17 at 21:15
  • @egreg Exactly. The limit of the numerator need not even exist. I posted This Answer to a soft question yesterday that discusses this point as one that is not often elucidated. And (+1) for this nice solution. -Mark – Mark Viola Feb 25 '17 at 04:05
  • @AlexM. Actually, the limit of the numerator need not even exist. But of course, it value must exist in a deleted neighborhood of the limiting value. – Mark Viola Feb 25 '17 at 04:08
  • This is an example where L'Hospital's Rule offers an easy and I say smart solution. But instead of such problems, the rule is mostly applied to trivial "differentiate and plug" type of problems. +1 – Paramanand Singh Feb 25 '17 at 04:16