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Define $L:(0,\infty)\to\mathbb{R}$ by


Show that $\lim_{x\to0^{+}}L(x)=-\infty$ and $\lim_{x\to\infty}L(x)=\infty$.

This is what I have done: We have that


So, it must be the case that $\lim_{x\to0^{+}}L(x)=\pm\infty$. Moreover, we have that


Hence, this implies that $\lim_{x\to0^{+}}L(x)=-\infty$.

For the other case, we have that


But since this sum does not converge, it must be the case that $\lim_{x\to\infty}L(x)=\infty$.

What do you guys think?

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Having a derivative with a vertical asymptote does not imply the same for the function. For example $f(x)=\sqrt{x}$ at $x=0$. The second part looks okay as long as you are not using the integral test to show that the series diverges. – Joe Johnson 126 May 1 '12 at 17:58
How do you get the harmonic series? Shouldn't the first term be $1/2$? – joriki May 1 '12 at 17:59
You need to prove the last inequality Josue. – Pedro Tamaroff May 1 '12 at 18:09
up vote 3 down vote accepted

You can show that $\lim_{x\to\infty}L(x)=\infty$ first, then deduce that $\lim_{x\to0^{+}}L(x)=-\infty$. But, as joriki mentioned, your estimate needs to be made more precise. As a hint:

$$ \int_{1}^{x} \frac{dt}{t} = \int_{\lfloor x \rfloor}^{x} \frac{dt}{t} + \sum_{k=1}^{\lfloor x \rfloor - 1} \int_{k}^{k+1} \frac{dt}{t}. $$

Then, once you've shown that $\lim_{x\to\infty}L(x)=\infty$, consider the substitution $t = 1/u$, giving

$$ \int_{1}^{x} \frac{dt}{t} = - \int_{1}^{\frac{1}{x}} \frac{du}{u}. $$

You only need to do the work once due to the symmetry of the function $1/t$. Integrating from $1$ to $\infty$ is basically the same as integrating from $0$ to $1$.

enter image description here

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I've gotten to the point where I need to show that $\int_{1}^{2}dt/t+\int_{2}^{3}dt/t+\int_{3}^{4}dt/t+\cdots=\infty$. But I'm having a hard time trying to formally derive an inequality to achieve this... – EsX_Raptor May 1 '12 at 19:23
Never mind: I just had an idea. Thanks! – EsX_Raptor May 1 '12 at 19:37
@JosuéMolina, You're welcome! What was your idea? – Antonio Vargas May 1 '12 at 19:41

The derivative of $f$ having an infinite limit does not imply that $f$ has an infinite limit. Consider $\lim\limits_{x\rightarrow0^+}\sqrt x$.

But, here's a hint:

You can show that $\int_\delta^{2\delta} {1\over t}\,dt\ge\delta\cdot{1\over2\delta}= {1/2}$ for any positive $\delta$. This would imply that $\int_0^1{1\over t}\,dt$ diverges to $\infty$ (consider the integral over the intervals intervals $[1/4,1/2]$, $[1/8,1/4]$, $[1/16,1/8]$, $\ldots\,$). From this, it follows that $\lim\limits_{x\rightarrow0^+}L(x)=-\infty$.

You could also show, by considering the integral over the intervals $[1 ,2 ]$, $[2,4]$, $[4,8]$, $\ldots\,$ that $L(2^n)\ge n\cdot {1\over2}$. This would imply that $\lim\limits_{x\rightarrow\infty}L(x)=\infty$.

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On your idea of proving that the sum $S(M)$

$$S(M)=\sum_{n=1}^M \int_n^{n+1}\frac {dt} t$$ you can argue as follows (make a graph, or use that $f$ is decreasing):

$$\frac{1}{n+1}<\int_n^{n+1}\frac {dt} t<\frac{1}{n}$$

$$\sum_{n=1}^M \frac{1}{n+1}<\sum_{n=1}^M \int_n^{n+1}\frac {dt} t<\sum_{n=1}^M \frac{1}{n}$$

$$\sum_{n=1}^M \frac{1}{n+1}<\int_1^{M+1}\frac {dt} t<\sum_{n=1}^M \frac{1}{n}$$ As you can see from other answers, the result for $\displaystyle \int_0^1 \frac{dt}t$ follows as a corollary.

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