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If $a > 0$ and $b > 0$, show that

$$\lim_{x \to +\infty}\frac{(\log x)^b}{x^a} = 0 \tag{1}$$

and

$$\lim_{x \to +\infty}\frac{x^b}{e^{ax}} = 0 \tag{2}$$


Attempts:

$(1)$

Given that

$$\log x = \frac{\ln x}{\ln 10}$$

Then $$ \begin{align*} \lim_{x \to +\infty}\frac{(\log x)^b}{x^a} &= \lim_{x \to +\infty}\left(\frac{\ln x}{\log 10}\right)^b\cdot \lim_{x \to +\infty}\frac{1}{x^a}\\ &= \lim_{x \to +\infty}\left(\frac{\ln x}{\log 10}\right)^b\cdot 0\\ &= 0 \end{align*} $$ I'm not sure whether this is right. There is a Theorem a my textbook which says:

If $f(x)$ is an infinitesimal function as $x \to a$, and $g(x)$ is a bounded function, then $\lim_{x \to a}f(x)\cdot g(x)$ is an infinitesimal (i.e $= 0$).

The rightmost limit it is indeed and infinitesimal function, but it seems that the leftmost one is unbounded. Does the theorem not hold here?

$(2)$ $$\lim_{x \to +\infty}\frac{x^b}{e^{ax}} = 0$$

Here I'm puzzled. So far I've done just this:

$$\lim_{x \to +\infty}\frac{x^b}{e^{ax}} = \lim_{x \to +\infty}\frac{e^{b \ln x}}{e^{ax}} = ...$$

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You must know something more, if you cannot use De l'Hospital. For instance $$ \lim_{x \to +\infty} \frac{x^b}{e^x}=0 \quad\hbox{for every $b>0$} $$ or $$ \lim_{x \to +\infty} \frac{x}{e^{ax}}=0 \quad\hbox{for every $a>0$}. $$ Indeed, $$ \frac{x^b}{e^{ax}}=\left( \frac{x}{e^{\frac{a}{b}x}} \right)^b $$ or $$ \frac{x^b}{e^{ax}}=\left( \frac{x^{\frac{b}{a}}}{e^{x}} \right)^a $$ For sure you cannot use the theorem in your book, since $\log x$ does not remain bounded as $x \to +\infty$.

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  • $\begingroup$ Well, I took both limits from the section treating polynomials approximations to functions. I haven't reached that part yet (I just wanted to see whether they were solvable by basic means). If the theorem does not hold there, what I have proved is not valid, right? Or it's valid but I need to apply another theorem? $\endgroup$ – Jazz Feb 25 '15 at 14:39
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    $\begingroup$ Your "proof" is wrong, since $[\infty \cdot 0]$ is an indeterminate form and it needn't be zero. $\endgroup$ – Siminore Feb 25 '15 at 14:48
  • $\begingroup$ Your answer assumes a result which is algebraically equivalent to the result we are trying to establish. A proper proof is presented in my answer. $\endgroup$ – Paramanand Singh Jul 25 '15 at 5:10
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This is a standard property of logarithmic and exponential functions and one result can be derived from another. We need the following things in order to prove it.

1) $\log x \leq x - 1$ for all $x \geq 1$.

2) $x^{k} \to \infty$ as $x \to \infty$ for $k > 0$.

Let $c = a/b$ so that we can write $$f(x) = \frac{(\log x)^{b}}{x^{a}} = \left(\frac{\log x}{x^{c}}\right)^{b} = \{g(x)\}^{b}$$ Now we will show that $g(x) \to 0$ as $x \to \infty$ and this will imply that $f(x) \to 0$ as $x \to \infty$.

Note that since $x \to \infty$ we can assume $x > 1$ and hence for any $d > 0$ we have $x^{d} > 1$ and therefore $$\log x = \log \{(x^{d})^{1/d}\} = \frac{\log x^{d}}{d} \leq \frac{x^{d} - 1}{d}$$ It is now obvious how to show that $g(x) \to 0$ as $x \to \infty$. We just need to take $d$ such that $0 < d < c$ so that $$0 < g(x) = \frac{\log x}{x^{c}} \leq \frac{x^{d} - 1}{dx^{c}} < \frac{x^{d}}{dx^{c}} = \frac{1}{d}\cdot\frac{1}{x^{c - d}}\tag{1}$$ Since $k = c - d > 0$ we have $x^{c - d} = x^{k} \to \infty$ and hence $1/x^{c - d} \to 0$. Applying Squeeze theorem to equation $(1)$ above when $x \to \infty$ leads us to the conclusion that $g(x) \to 0$ as $x \to \infty$. And then $f(x) = \{g(x)\}^{b} \to 0$ as $x \to \infty$.

Note that the accepted answer by Siminore uses an algebraical equivalent of the result which it is trying to prove so it is kind of circular.

Handling the limit of $h(x) = x^{b}/e^{ax}$ is simple if we put $y = e^{x}$ so that $x = \log y$ and then $y \to \infty$ as $x \to \infty$ and our function $h(x) = (\log y)^{b}/y^{a} \to 0$ (from previous part of this answer).

Note: I would like to mention few points about your post. When you are studying calculus/analysis then $\log$ is same as $\ln$ and base $10$ logarithms are written as $\log_{10}x$ explicitly. So you don't need to think along the lines of conversion between common logs and natural logs.

Another point is about the theorem which you mention. The use of word "infinitesimal function" adds so much confusion here. It is better to rewrite the theorem as follows:

Theorem: If $f(x) \to 0$ as $x \to a$ and $g(x)$ is bounded in a deleted neighborhood of $a$ then $f(x)g(x) \to 0$ as $x \to a$.

Also as you have noted this theorem is of no use in solving this problem.

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2: Assuming a and b are both positive and finite (second part wasn't mentioned, but I am assuming it). Expand the exponential term in the denominator as a sum.

$\frac{x^b}{e^{ax}}=\frac{x^b}{1+ax+\frac{1}{2}(ax)^2+\frac{1}{6}(ax)^3+....+\frac{1}{n!}(ax)^n}$ for $n\rightarrow\infty$. You can now use L'Hopital's rule on this in the limit where $x\rightarrow\infty$. In particular you will have some n that is large enough to cancel the $x^b$ term in the numerator, and still have some larger powers of x in the denominator. (e.g. you will be able to use L'Hopital's rule b times and still have a term dependent on x in the denominator.) Because $x\rightarrow\infty$, we are guaranteed that $(ax)>1$ for any finite $a$.

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