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I want to compute the following limit: $$\lim_{x\to 0}\frac{\int_0^x {e^{t^5+1}-5} dt }{x}$$ I have thought to use the de l'Hopital rule, but in order to be able to apply this result I have to verify:

  1. if the limit gives us an indeterminate form of kind $\frac{0}{0}$ or $\frac{\infty}{\infty}$
  2. if the ratio of the derivative of the functions at the numerator and denominator exists.

I want to prove $\textbf{1.}$

Surely the limit of the denominator gives me $0$, so necessarily I want that the numerator goes to $0$.

$\textbf{First way:}$ I have thought that trivially, since when $x\to 0$ I would have $\lim_{x\to 0}\int_0^x {e^{t^5+1}-5} dt \to \int_0^0 {e^{t^5+1}-5} dt $ and for the fact that in this last integral the upper and lower limits of the integral coincide and they are equal to 0, I have that the limit is $0$ (I am using the fact that in general $\int_a^a f(t)dt=0$).

$\color{red}{\text{First doubt:}}$

$\textbf{I am not conviced of this fact...if my idea is right this would imply that}$ $\textbf{ in general whatever function of kind $\int_0^x f(t) dt$ goes to $0$ when $x\to 0$}$. This could be a conseguence of the fact that whatever function $G(x)=\int_0^x f(t) dt$ is continous and so $\lim_{x\to 0}G(x)=G(0)=0$...am I right or I am failing somewhere?

Alternatively I have also thought two different ways to prove that the limit of the numerator is $0$.

$\textbf{First alternative way:}$ The function $f(t)=e^{t^5+1}-5$ is continous and differentiable $\forall t\in \mathbb{R}$ so I can apply the mean value integral theorem in whatever $[0,x]$ with $x>0$ and so: $$\exists \xi\in[0,x]: \int_{0}^x e^{t^5+1}-5=f(\xi)\cdot (x-0)=(e^{\xi^5+1}-5)\cdot x$$ So $\lim_{x\to 0}\int_{0}^x (e^{t^5+1}-5)=\lim_{x\to 0}(e^{\xi^5+1}\-5)cdot x=0$ (remembering that when $x\to 0$ also $\xi\to 0$)

$\textbf{Second alternative way:}$ The function $f(t)=e^{t^5+1}-5$ is continous $\forall t$ and in particular for $t=0$ so $\forall a>0$ I can say: $$\text{in }[0,a] \exists m,M\in\mathbb{R}: m\leq f(t)\leq M$$ This holds from Weierstrass because in $[0,a]$ I am in a neighbourhood of $0$, and in $0$ the function is continous, and since I want to understand what happens near $0$ (because I consider $x\to 0$) I can say: $$0=\lim_{x\to 0}mx=\lim_{x\to 0}\int_0^xmdt\leq\lim_{x\to 0}\int_0^x {e^{t^5+1}-5} dt \leq \lim_{x\to 0}\int_0^xMdt=\lim_{x\to 0}Mx=0$$ $\color{red}{\text{Second doubt:}}$

I am not so convinced of these tw different ways of proving the limit. Do you thinks they are correct? If yes what is in your opinion the best way to use (the first or one of the last alternative)?

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    $\begingroup$ Continuity of the indefinite integral does tell you that the numerator tends to $0$. You are just ovethinking. $\endgroup$ Apr 19, 2021 at 7:19
  • $\begingroup$ So my first idea is correct? $\endgroup$
    – pawel
    Apr 19, 2021 at 7:20
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    $\begingroup$ Yes, that is what I am saying. $\endgroup$ Apr 19, 2021 at 7:21
  • $\begingroup$ Ok thanks! And the other two ways are correct too? $\endgroup$
    – pawel
    Apr 19, 2021 at 7:43
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    $\begingroup$ If $f$ is any function Riemann integrable on $[a, b] $ then $F(x) =\int_a^x f(t) \, dt$ is continuous on $[a, b] $. And hence the limit of your numerator is $0$. $\endgroup$
    – Paramanand Singh
    Apr 19, 2021 at 9:43

2 Answers 2

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If a function $f$ is continuous over some interval $I$, then the integral function $$ F(x)=\int_a^x f(t)\,dt $$ is (defined and) differentiable over $I$. Here $a\in I$ is arbitrary. This is the fundamental theorem of calculus.

In particular $F$ is continuous over $I$. So, in your case, you don't even have to compute the limit of the numerator, exactly like you don't need to compute the limit of the denominator, in order to check whether the assumptions of l'Hôpital's theorem are satisfied.

By the way, you're missing one of the main assumptions: the denominator should be differentiable in a punctured neighborhood of $c$, with nonzero derivative in this neighborhood (talking about limits for $x\to c$).

About your “alternative” ways: how do you prove the integral mean value theorem without having proved the existence of the integral function? And you're stating it wrongly. You can say that, if $f$ is a continuous function over $I$ and $a,b\in I$, then there exists $c\in (a,b)$ such that $$ (b-a)f(c)=\int_a^b f(t)\,dt $$ Not $f'(c)$. And the derivative of $t\mapsto e^{t^2+1}$ is $2te^{t^2+1}$ (but you don't need this derivative).

Final remark: use parentheses and write $$ \int_0^x(e^{t^2+1}-5)\,dt $$

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  • $\begingroup$ Thanks! some clarifications: 1) So can I say that the limit is zero simply since I will obtain an integral with upper and lowe bounds equal? 2) In the application of the mean value theorem there was a typo, obviuosly it was $f(\xi)$ and not $f'(\xi)$. Moreover the existence of the integral function follows trivially by the continuity of the function $e^{t^2+1}-5$, not? 3) the third way do you think is correct? $\endgroup$
    – pawel
    Apr 19, 2021 at 9:48
  • $\begingroup$ @YvesDaoust when I compute the limit for the fact that the integral function (G(x)) is continuous I can say that $G(x)\to G(0)=\int_0^0$, not? $\endgroup$
    – pawel
    Apr 19, 2021 at 10:01
  • $\begingroup$ @YvesDaoust Yes but the limit coincides with the value of the function in x=0 by the continuity! $\endgroup$
    – pawel
    Apr 19, 2021 at 10:31
  • $\begingroup$ @YvesDaoust I think you have not understood what i have saying...I am saying that by continuity the numerator tends to 0, and also the denominator tends to $0$ and so then I will apply de l'Hopital! I stress the fact that I am not saying without applying de l'Hopital that the limit is zero but I am saying that the integral function goes to $0$ by continuity! $\endgroup$
    – pawel
    Apr 19, 2021 at 11:19
  • $\begingroup$ @YvesDaoust: I think all the comments of asker are trying to say that limit of numerator is $0$. Even the question is about checking their approach of proving limit of numerator is $0$. The question is not about evaluating the limit of the fraction given in question. $\endgroup$
    – Paramanand Singh
    Apr 19, 2021 at 11:42
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Let $f(t) =e^{t^5+1}-5$. Let us observe that $f$ is continuous everywhere so that the function $F$ given by $F(x) =\int_0^x f(t) \, dt$ is defined everywhere. Note also that functions defined as integrals of Riemann integrable functions are continuous (proof later) and hence $F$ is continuous everywhere.

Thus $F(x) \to F(0)=0$ as $x\to 0$. Your question was mostly about establishing this.

Next we come to the limit of the fraction in question. We have to evaluate $\lim_{x\to 0}F(x)/x$. If you observe carefully this limit is by definition the derivative $F'(0)$ (write the definition of derivative and confirm this).

Next we use fundamental theorem of calculus which says that if $f$ is continuous at some point $c$ then $F'(c) =f(c) $. And thus $F'(0)=f(0)=e-5$.


Now to the proof of continuity of function defined by integrals. Let $f:[a, b] \to\mathbb {R} $ be Riemann integrable on $[a, b] $. By definition of Riemann integral the function $f$ is bounded on $[a, b] $ and hence we have a number $M>0$ such that $|f(x) |<M$ for all $x\in[a, b] $.

Let $F(x) =\int_{a} ^x f(t) \, dt$ and consider a point $c\in [a, b] $ and another point $x\in [a, b] $ and we have $$|F(x) - F(c) |=\left|\int_c^x f(t) \, dt\right|\leq\left|\int_c^x|f(t)|\, dt\right|\leq M|x-c|$$ It follows by definition of limit (you can choose $\delta=\epsilon/M$ in limit definition) that $F(x) \to F(c) $ as $x\to c$. This proves the continuity of $F$ at $c$. Since $c$ was an arbitrary point of interval $[a, b] $ it follows that $F$ is continuous on $[a, b] $.

Compare this proof with your second alternative approach.

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  • $\begingroup$ Thanks really a lot! The last question: the fact that $\lim_{x \to 0}\frac{F(x)}{ x}$ is the derivative of $F(x)$ in $0$ follows from the fact that I have to write the definition of derivative in $x_0$ as $\lim_{x\to x_0}\frac{F(x)-F(x_0}{x-x_0}$? $\endgroup$
    – pawel
    Apr 19, 2021 at 15:52
  • $\begingroup$ @pawel: yes replace $x_0$ by $0$ because you need derivative at $0$ and note that $F(0)=0$. $\endgroup$
    – Paramanand Singh
    Apr 19, 2021 at 16:18
  • $\begingroup$ Perfect! Again thanks a lot for confirming my ideas and for this additional proof! $\endgroup$
    – pawel
    Apr 19, 2021 at 16:26

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