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Let $\,\,f:\mathbb{R}\rightarrow\mathbb{R}\,\,$ be a continuous function such that $ \ \int_{0}^{\infty}{f(x)} dx $ exists. Which of the following statements are always true ?

1.if $\lim_{n\rightarrow\infty}f(x)$ exists, then $\lim_{n\rightarrow\infty}f(x)=0$
2.$\lim_{n\rightarrow\infty}f(x)$ exists and is $0$
3.in case $f$ is nonnegative function ,$\lim_{n\rightarrow\infty}f(x)$ must exist and zero
4.in case of $f$ is differentiable function , $\lim_{n\rightarrow\infty}$f$'$(x) must exist and is zero

I think 1 is always true , no idea for other three
please help

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I guess you mean the limit when $x\to\infty$. –  Hui Yu Oct 29 '12 at 7:45

2 Answers 2

up vote 3 down vote accepted

The assertion (2) is not always true. For any positive integer $n$, let $f(n-\frac{1}{2^n})=f(n+\frac{1}{2^n})=0$, and let $f(n)=1$.

In the interval $[n-\frac{1}{2^n},n]$, let $f(x)$ climb linearly to $1$, and in the interval $[n,n+\frac{1}{2^n}]$, let $f(x)$ fall linearly to $0$. For every other $x\ge 0$, let $f(x)=0$.

The integral of $f(x)$ in the interval $[n-\frac{1}{2^n},n+\frac{1}{2^n}]$ is just the area $\frac{1}{2^n}$ of a triangle. The sum of these areas is finite. So the improper integral exists, but the function does not have limit $0$.

Our function $f(x)$ is non-negative, so it also provides a counterexample to assertion (3).

We can smooth out the function described above and obtain a counterexample to (4). But (4) has many other types of counterexamples. In general, even if we know that $f(x)$ is very close to $0$, we can say very little about $f'(x)$.

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You're right. Think - what would have happened if $$\exists a\ s.t.\lim_{n\rightarrow\infty}f(x)=a>0$$

In that case, there's a positive M such that for any $x>M$, $f(x)>a$. (Deal with the negative case similarly). You bound the function away from zero for any x>M. Then the integral can be written as: $$\int_0^\infty f(x)dx = \int_0^M f(x)dx +\int_M^\infty f(x)dx $$ and the third integral in this equation is infinite.

no. 2 has some simple counter examples. (take a function with a narrowing sequence of "spikes").

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Not sure to get the hint. (And please avoid this mishmash of maths symbols and English.) –  Did Oct 29 '12 at 7:54

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