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Like I said I need to find the limit

$\lim_{x\to 0} \frac{1}{x} \cdot \int_{0}^{x}(1+\sin(t))^\frac{1}{t}dt$

I think I have to use L'Hospital here since the test will be about it but I can't understand how to solve this exercise at all.

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  • $\begingroup$ Good idea. L'Hospital's rule requires you to differentiate the numerator and denominator of some fraction. Which fraction would you like to take, and can you differentiate it? $\endgroup$
    – B. Mehta
    Nov 28, 2017 at 21:48
  • $\begingroup$ The problem I have is showing that I can even use L'Hospital with the integral that I have. I would first have to prove something with the integral but what? $\endgroup$
    – John.Doh
    Nov 28, 2017 at 21:54
  • $\begingroup$ What are the conditions to apply L'Hospital? $\endgroup$
    – B. Mehta
    Nov 28, 2017 at 21:56
  • $\begingroup$ either $\frac{0}{0}$ or $\frac{\inf}{\inf} $ $\endgroup$
    – John.Doh
    Nov 28, 2017 at 21:59
  • $\begingroup$ Right, which case do you have here? $\endgroup$
    – B. Mehta
    Nov 28, 2017 at 21:59

3 Answers 3

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You do indeed have a quotient, so L'Hôpital does apply. Specifically, the limit is $\lim_{x \to \infty} \frac{\int_{0}^{x}(1+sin(t))^\frac{1}{t}}{x}$.

To actually invoke L'hopital, check that your form is indeed indeterminate (i.e. 0/0, $\infty/\infty$, etc). The bottom is obviously infinity. The top is a little less obvious. Does this definite integral expression keep growing as x keeps growing?

The answer is yes. The integrand is always positive after a while, so you won't have cancellation issues. Additionally, because raising something to $\frac{1}{t}$ for large t is essentially just taking a giant nth root (what happens when you take giant nth roots of numbers greater than 1? Positive numbers less than 1?), the integrand hovers around f(x)=1 for large x. That means your integral (area-finding) expression is going to go off to infinity (the infinite integral of a roughly constant function diverges).

Great, so we have $\infty/\infty$. This is an indeterminate form, and L'hopital's rule applies. Now differentiate the top and bottom.

Differentiating the bottom is easy (it's just $1$). Differentiating the top is also easy (remember fundamental theorem of calculus?)—it's just the integrand since we're going from 0 to x.

Your limit expression is now just

$\lim_{x \to \infty} \frac{(1+sin(t))^\frac{1}{t}}{1}$.

EDIT: IGNORE EVERYTHING BENEATH THIS LINE. Apparently the limit was approaching 0, not $\infty$, so we actually avoid the weird properties I discuss below.

Common sense should tell you the limit is 1, because no matter what we're taking the "giant nth root" of a positive number, and that always goes to 1. We should prove it rigorously though. (If we prove this rigorously, then we also prove that the integral of the expression diverges as x goes to infinity. Can you reason out why? Think in terms of epsilons and deltas, or perhaps greatest upper bounds.)

Break this down case-by-case. Since the thing inside the parentheses oscillates between 0 and 2, and each of those values in between gets raised to an extremely small power, consider the limits of all non-zero quantities as $\frac{1}{t}$ goes to zero. Because you know something about "giant nth roots", you should be able to reason that these limits are all 1.

What about the weird case when the thing in parentheses is zero? We have to prove that as we approach zero, maintaining a high power, those things stay near 1. (We're really proving that the points where the function dips $\epsilon$-away from 1 form a set of arbitrarily small measure after some x > N, which is the best we're gonna get, since 0 to any finite power vanishes).

We can simply argue this by saying "let the parenthetical term be any $\epsilon > 0$. Then for all $\epsilon > 0$, there's a small enough $1/t$ such that we get really close to 1."

This problem is a bit weird because of the property that the function dips to exactly zero once every $2\pi$ units, but the above argumentation (for the purposes of your class I'm not sure how necessary or lacking it is) should be enough to prove that the desired limit is 1.

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  • $\begingroup$ Like I mentioned before I wrote $x \to \inf$ however it should have been $x \to 0$ $\endgroup$
    – John.Doh
    Nov 28, 2017 at 22:06
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$$\lim_{x\rightarrow\infty}\frac{\int_{0}^{x}(1+\sin{t})^{1/t}dt}{x}=\lim_{x\rightarrow\infty}\frac{d}{dx}\int_{0}^{x}(1+\sin{t})^{1/t}dt=$$ $$=\lim_{x\rightarrow\infty}(1+\sin{x})^{1/x}=1$$

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  • $\begingroup$ Aaaaah I am so sorry, I wrote $x \to \inf$ however it should have been $x \to 0$. does that change anything? $\endgroup$
    – John.Doh
    Nov 28, 2017 at 22:02
  • $\begingroup$ @John.Doh The difference is that everywhere $\lim_{x\to\infty}$ is replaced with $\lim_{x\to0}$, which changes the final answer. To compute $\lim_{x\to0}(1+\sin x)^{1/x}$, you need to use l'Hopital's rule again. First, compute the limit of its natural log, $\lim_{x\to0}\ln((1+\sin x)^{1/x})$, which can be done using l'Hoptial's. The final answer is e to this result. $\endgroup$ Nov 28, 2017 at 22:07
  • $\begingroup$ what rule do you use to do the first equation? $\endgroup$
    – John.Doh
    Nov 28, 2017 at 22:19
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If it is indeed the limit when $x\to 0$, the limit of this quotient is but the definition of the derivative at $0$ of the function $$\begin{cases}f(x)=\displaystyle\int_0^x(1+\sin t)^{\tfrac1t}\mathop{}\!\mathrm d t &\text{if } x\ne 0\\[1ex]f(0)=0\end{cases}$$ By the First fundamental theorem of integral calculus, $$f'(x)=(1+\sin x)^{\tfrac1x} \quad (x\ne 0).$$ and $\;f'(0)=\lim_{x\to 0}f'(x)$ if this limit exists.

So all we have to do is finding this limit. We'll find first the limit of its logarithm: $$\ln f'(x)=\frac1x\ln(1+\sin x)\sim_0\frac{\sin x}x\to 1, $$ whence $$\lim_{x\to 0} f'(x)=\mathrm e =f'(0). $$

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