Let $f$ be a continuous function on $[0,\infty)$ such that $\lim_{x\to \infty}(f(x)+\int_0^x f(t)dt)$ exists. Find $\lim_{x\to \infty}f(x)$.
Useful hints will work. Please give some nice hint to solve the problem.
Let $f$ be a continuous function on $[0,\infty)$ such that $\lim_{x\to \infty}(f(x)+\int_0^x f(t)dt)$ exists. Find $\lim_{x\to \infty}f(x)$.
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$\begingroup$
integration
limits
functions
continuity
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0Try taking $g=f+K$ with $K$ some constant. What happends? Can you guess the value of the limit? – 2017-02-02
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0@xavierm02 taking $g$=$f+K$ we will get $\lim_{x\to \infty} f(x)+K$=$\lim_{x\to \infty}g(x)$.Is it correct ? If yes whats next? – 2017-02-02
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0But what is $\lim_{x\to \infty}(g(x)+\int_0^x g(t)dt)$? – 2017-02-02
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0I didn't get that.Did you mean $\lim_{x\to \infty}(f(x)+\int_0^x f(t)dt)$? – 2017-02-02
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0No I meant with $g$, under the assumption that the limit exists with $f$. – 2017-02-02
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0If we assume that $\lim_{x\to\infty}f(x)$ exists and is greater than $0$, it follows that $\lim_{x\to\infty}\int_0^xf(t)\ dt$ exists, is non-zero, and of the opposite sign. That is, $\int_0^\infty f(t)\ dt$ exists. But if $\lim_{x\to\infty}f(x)$ exists, it follows that there must exist $x'$ for every $\epsilon>0$ such that for all $x>x'$, we have $|L-f(x)|<\epsilon$. Can you take it from here? – 2017-02-02
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0@SimplyBeautifulArt How we get to know that $\lim_{x\to\infty}\int_0^xf(t)\ dt$ is of opposite sign ? please can you explain it in simpler terms ? – 2017-02-02
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0@xavierm02 plz help me in proceeding further.I am not getting the logic behind defining a new function $g$. – 2017-02-03
2 Answers
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One of the easiest approaches is the use of L'Hospital's Rule here. Let $F(x) =\int_{0}^{x}f(t)\,dt$ so that $F'(x) =f(x) $ via Fundamental Theorem of Calculus. We are thus given that $F(x) +F'(x) \to L$ as $x\to\infty$.
We have \begin{align} \lim_{x\to\infty} F(x) &=\lim_{x\to\infty} \frac{e^xF(x)} {e^x}\notag\\ &=\lim_{x\to \infty} \dfrac{e^x\{F(x) +F'(x) \}} {e^x}\text{ (via L'Hospital's Rule)} \notag\\ &=\lim_{x\to\infty} F(x) +F'(x)\notag\\ &=L\notag \end{align} and hence $$\lim_{x\to \infty} f(x) =\lim_{x\to\infty} F'(x) =\lim_{x\to\infty} \{F(x) + F'(x) \}-F(x) =L-L=0$$ Based on feedback received in comments, it appears that some more explanation of the steps is needed.
Note that the given hypotheses imply that limit of $$f(x) +F(x) =F'(x) +F(x) =\frac{e^x(F(x) +F'(x)) } {e^ x}=\frac{(e^xF( x)) '} {(e^x)'} $$ exists and equals $L$ as $x\to \infty $. By L'Hospital's Rule it follows that the limit of $\dfrac{e^xF(x)} {e^x} $ (which is same as $F(x) $) also exists and equals $L$.
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1Nice solution, +1. But the flow is weird because you write $\lim$ when it hasn't been shown the limit exists. Also, you don't know the limit of the numerator, $e^xF(x),$ is $\infty.$ Now to use L'Hopital in this case, you don't need to know that. But many readers may not know this little fact. – 2018-11-30
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0@zhw : I should have added more details. Thanks for putting these details in your comment. – 2018-11-30
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Let $F(x) = \int_0^x f(t)dt$. Then, we have $\lim_{x\to \infty} e^{-x}[e^x F(x)]' = c$ exists. From this, given $\epsilon>0$, there is $N>0$ such that for all $x>N$, $$ (c-\epsilon)e^x \leq[e^x F(x)]' \leq (c+\epsilon)e^x. $$ Now, integrating from $N$ to $t$ gives: $$(c-\epsilon)(e^t-e^N)\leq e^tF(t) -e^NF(N) \leq (c+\epsilon)(e^t-e^N),\quad\forall t>N $$and hence $$ (c-\epsilon)(1-e^{N-t})\leq F(t) -e^{N-t}F(N) \leq (c+\epsilon)(1-e^{N-t}), \quad \forall t>N. $$Taking $t\to\infty$ yields $$c-\epsilon \leq \liminf_{t\to \infty}F(t)\leq \limsup_{t\to \infty}F(t)\leq c+\epsilon. $$Since $\epsilon>0$ was arbitrary, we have $\lim_{t\to\infty} F(t) =c$ and thus $\lim_{t\to\infty} f(t) = 0$.