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I have a lot of trouble wrapping my head around using Squeeze Theorem to find the limits (mostly of sequences of integrals, as these are the exercises that I want to get better at right now).

I have asked a lot of questions regarding such limits so far, but I really want to get some insight as to how you actually go about solving them. I have a few examples that we can go through.

First, finding the $\lim_{n\to\infty}I_n,\,(I_n)_{n\ge1},\, I_n = \int_0^1{\frac{x^n}{x^2+2x+3}\,dx}$. A general approach that I go with is $1)$ check if $I_n$ is increased\decreasing then find its bounds by considering the integral bounds and rewriting from there to resemble a double inequality with $I_n$ in the middle. For this example $I_n$ is decreasing and $0 \le I_n \le \frac{1}{3}$. This helps me understand that $I_n$ is convergent but the bounds won't help with using Squeeze Theorem. Next $2)$ I try to find a function that can be substituted as the bound and also whose limit when integrated from $0$ to $1$ in this case will yield the same result as the other bound of the $I_n$. Now after practicing a lot I can easily say that $x^n$ would be a good upper bound because the value under integral is $x^n$ divided by a number greater than 1, but also because $\int_0^1x^n \,dx = \frac{1}{n+1}$ and the limit of this will tend to $0$ as does the lower bound$(0)$. So this is pretty straightforward for me now.

What about $\lim_{n\to\infty}I_n, \, I_n=\int_0^1\frac{\ln(x^n+1)}{x+1} \,dx$. $1) I_n$ is decreasing, $0 \le I_n \le \frac{\ln^22}{2}$. So right now I am thinking since $I_n$ is decreasing with a maximum of $I_1=\frac{\ln^22}{2}$ that maybe the limit will tend to $0$. $2)$ Now I would need to find a function whose limit after being integrated from $0$ to $1$ will be equal to $0$. So I am thinking about the value under the integral, it is a fraction. The denominator $x+1 \ge 1$ so I realize that removing it would guarantee that the value of the function under the integral will increase or stay equal for $x=0$, so it is an upper bound. Then I can certainly say $0 \le \frac{\ln(x^n+1)}{x+1} \le \ln(x^n+1)$. When integrating all sides I get $ 0 \le I_n \le \int_0^1\ln(x^n+1)$. I can try calculating the value of the new integral now but I don't even know how to (I tried integrating by parts without very much success) and I have no guarantee the limit of its value will tend to $0$ anyway.

I am really sorry for the lengthy post but I really need some general tips so I can understand how to solve this kind of exercises. I have asked questions about particular exercises before but I rarely get to actually understand how to solve them by myself afterwards.

Radu Gabriel
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    Do you know the inequality $\log(1+t)\leq t$? It is a very fundamental property of logarithms. Then your integral is less than $\int_0^1 x^n, dx$ and you are done. – Paramanand Singh Mar 18 '20 at 02:01
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    You will find that such problems usually also require familiarity with basic inequalities satisfied by functions under consideration. Sometimes things can be really tricky. – Paramanand Singh Mar 18 '20 at 02:03
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    In such problems evaluating integrals should be the last option. – Paramanand Singh Mar 18 '20 at 02:04
  • @Paramanand Singh. Honestly I realized $\log (x^n+1) \le x^n $ and realized that this would solve the second example just fine but it felt wrong at the time, I did not know this is a fundamental property of logarithms, I just thought of it and quickly dismissed it. Maybe these problems are not that hard after all. Thanks for your help, I really appreciate it. – Radu Gabriel Mar 18 '20 at 07:37
  • That inequality is fundamental and defining characteristic of natural logarithms. See this thread. – Paramanand Singh Mar 18 '20 at 08:55

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