# LHôpital's rule: fundamental theorem of calculus when the upper limit is a definite integral

I am just starting to learn Calculus and we're at limits/derivatives and integrals. I need to apply L'Hôpital's rule and I am using the fundamental theorem of calculus. I need to solve this example: $$\lim_{x\to 1} \frac{ \int_0^{\int _{1} ^ x (ue^{-u})du} \, (t^2e^t ) \, dt}{(x^3-1)}$$

The main instruction says to solve the limit using L'Hôpital's rule. But, I don't know exactly how to apply L'Hôpital's rule when the upper limit is a definite integral.

I tried it in this way

$$\lim_{x\to 1} \frac{ \frac{d}{dx} (\int_0^{\int _{1} ^ x (ue^{-u})du} \, (t^2e^t ) \, dt)}{ \frac{d}{dx} (x^3-1)}$$

So I get $$\lim_{x\to 1} \frac{(\int_1^x ue^{-u}du)^2 e^{\int_1^x ue^{-u}du}}{3x^2 } \,$$ I think that I am wrong in the way of interpreting the fundamental theorem of calculus part 1. What do you think?. Thank you.

• Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Feb 9 at 9:25
• You can use MathJax to format mathematics on this site. Enclose your formatting with dollar signs: \$x^2\$ becomes $x^2$. – Toby Mak Feb 9 at 9:57
• Thanks. I understand. – Charlie Van Basten Øydne Feb 9 at 9:58
• I tried to edit that but it seems you are just trying to make it such that no one can read it – user715522 Feb 9 at 10:14
• Done. I got it. I dont used to code in mathjax but it's done now. – Charlie Van Basten Øydne Feb 9 at 10:22

Let $$f(x) =\int_{1}^{x}ue^{-u}\,du$$ so that $$f(x)\to f(1)=0$$ as $$x\to 1$$ and by Fundamental Theorem of Calculus we have $$f'(1)=1e^{-1}=1/e$$.
Next note that the denominator of the expression under limit can be factorized as $$(x-1)(x^2+x+1)$$ and hence the desired limit equals $$\frac{1}{3}\lim_{x\to 1}\frac{\int_{0}^{f(x)}t^2e^t\,dt}{x-1}$$ which can be expressed further as $$\frac{1}{3}\lim_{x\to 1}\frac{f(x)-f(1)}{x-1}\cdot \frac{1}{f(x)}\int_{0}^{f(x)}t^2e^{t}\,dt$$ (note that $$f(1)=0$$).
The first factor under limit tends to $$f'(1)=1/e$$ via definition of derivative and the second factor tends to $$\lim_{u\to 0}\frac {1}{u}\int_{0}^{u}t^2e^t\,dt$$ via substitution $$u=f(x)$$. And by Fundamental Theorem of Calculus the above limit equals $$0^2e^0=0$$. Thus the desired limit is $$0$$.
You can see that directly evaluating the numerator and denominator at $$x=1$$ leads to the indeterminate form $$0/0$$. By L'Hôpital, $$\begin{split} L&=\lim_{x\to 1} \frac{\int_{0}^{\int_{1}^{x}\left(u e^{-u}\right) d u}\left(t^{2} e^{t}\right) d t}{\left(x^{3}-1\right)} = \lim_{x \to 1} \frac{\frac{d}{dx} \int_{0}^{\int_{1}^{x}\left(u e^{-u}\right) d u}\left(t^{2} e^{t}\right) d t}{\frac{d}{dx}\left(x^{3}-1\right)} \\ &= \lim_{x\to 1} \frac{\left[\left(\int_{1}^{x}\left(u e^{-u}\right) d u\right)^2 e^{\left(\int_{1}^{x}\left(u e^{-u}\right) d u\right)} \right] \cdot \frac{d}{dx}\left(\int_{1}^{x}\left(u e^{-u}\right) d u\right)}{3x^2} \\ &=\lim_{x\to 1} \frac{[(0) e^0 ] \cdot (1)e^{-1} }{3(1)^2} = 0, \end{split}$$ where we have used the chain rule, $$\frac{d}{dx}\Bigg|_{x=1} \int_0^{g(x)} t^2e^t dt = \left[\frac{d}{dy}\Bigg|_{y=g(1)} \int_0^yt^2e^tdt \right] \cdot \frac{d}{dx}\Bigg|_{x=1} g(x),$$ and the Fundamental Theorem of Calculus, $$\frac{d}{dx} \int_a^x j(t)dt=j(x).$$