Parametric limit with sine

Question

I'm trying to solve this limit problem which asks to find what happens to the following limits as $x$ varies:

$$\lim_{n\to+\infty} \left((\sin x -1) + {1\over{n^2 +1}}\right)^{n^2}$$

My steps so far: use exponential rule and rewrite as

$$\lim_{n\to +\infty} \exp\left(n^2 \ln \frac{n^2\sin x+\sin x-n^2}{n^2+1}\right)$$ Divide both the numerator and denominator in the argument of the logarithm by $n^2$ to get $$\frac{\sin x+\frac{\sin x}{n^2}-1}{1+\frac{1}{n^2}}$$

Then, as $n \to +\infty$ everything should behave like:

$$\lim_{n\to +\infty} e^{{n^2} \ln(\sin x -1)}$$

Hence we should have:

$$\lim_{n\to +\infty} (\sin x -1)^{n^2}$$

Answer 1

Let $q = \sin(x)-1$. The range of $q$ is $[-2, 0]$. There are three cases: $-2 \leq q < -1$, $q = -1$, and $-1<q\leq0$.

If $-2 \leq q < -1$, the limit does not exist, as $q^{n^2}$ oscillates between $\pm \infty$ as $n\to\infty$.

If $q = -1$, the limit also does not exist, as $(-1)^{n^2}$ oscillates between $\pm 1$.

If $-1 < q \leq 0$, the limit is $0$, as $q^{n^2}$ will approach $0$.

Answer 2

I'll expand on the answer of autimaticallyGenerated, because it has the right ideas but may be a bit short on explanation. Considering the intervals where $q:=\sin(x)-1$ lays is important, because, as you both realized, the question is very similar, but not identical, to considering

$$\lim_{n\to\infty}q^{n^2}. \label{eq1}\tag{1}$$

It's not identical because the real problem

$$\lim_{n\to\infty}\left(q+\frac1{n^2+1}\right)^{n^2} \label{eq2}\tag{2}$$

has that additional term. Let's consider \eqref{eq1} first, because it is a fundamental sequence, used very often.

If $\lvert q\rvert < 1$, then $\lvert q^n\rvert$ will become smaller and smaller as $n$ increases, so we get $\lim_{n\to\infty}q^{n^2}=0$ in this case (to get $\lvert q^n\rvert < \epsilon$, one needs to choose $ n > \frac{\ln \epsilon}{\ln \lvert q \rvert}$).

If $\lvert q\rvert > 1$, then $\lvert q^n\rvert$ will become bigger and as $n$ increases, and we have $\lim_{n\to\infty}|q|^{n^2}=+\infty$ in this case (to get $\lvert q^n\rvert > M$, one needs to choose $ n > \frac{\ln M}{\ln \lvert q \rvert}$). If $q$ is positive, that means $\lim_{n\to\infty}q^{n^2}=+\infty$, if $q$ is negative, that means the absolute value of $q^{n^2}$ will increase above any bound, but the sign will change in each step, as $n^2$ changes between odd and even in each step. So the sequences does not converge in that case.

If $q=1$, the sequence is constant and converges to $1$. If $q=-1$, the sequence is alternating between $+1$ and $-1$, so is not converging.

Make sure you really understand each case, as it will be helpful for the 'real' problem \eqref{eq2}.

Obviously we have

$$\lim_{n\to\infty}\left(q+\frac1{n^2+1}\right)=q, \tag{3}\label{eq3}$$

so it is very natural to try to use the same intervals for $q$ that have been useful ins solving \eqref{eq1}.

Let's turn to \eqref{eq2} now and consider $-1 < q \le 0$ first (the upper half of the possible interval for $q$). $q=0$ is trivial, the limit is $0$ in this case as $\left(\frac1{n^2+1}\right) \le \frac12, \forall n \ge 1$, so $\left(\frac1{n^2+1}\right)^{n^2} \le \left(\frac12\right)^{n^2}, \forall n \ge 1$ and we know from \eqref{eq1} that the right hand side converges to $0$.

If$-1<q<0$, then there is some $n_0$ such that $-1 < q < q+\frac1{n^2+1} < 0, \forall n \ge n_0$. That means we have $\left\lvert\left(q+\frac1{n^2+1}\right)^{n^2}\right\rvert < \left\lvert q^{n^2}\right\rvert, \forall n \ge n_0$. We know that the term on the right hand side goes to $0$ for $n\to\infty$ from the considerations for \eqref{eq1}. So we get our first partial result

$$\lim_{n\to\infty}\left(q+\frac1{n^2+1}\right)^{n^2}=0, \text{ if } -1 < q \le 0. $$

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Let's postpone $q=-1$ for the moment, and let's consider $-2 \le q < -1$ first. This means we can find a $q'$ with $q < q' < -1$. We also know from \eqref{eq3} that there is some $n_0$ with $\left(q+\frac1{n^2+1}\right) < q', \forall n \ge n_0$. That means that we have

$$\left\lvert\left(q+\frac1{n^2+1}\right)^{n^2}\right\rvert > \lvert q'\rvert^{n^2}, \forall n \ge n_0.$$

But we know how $\lvert q'\rvert^{n^2}$ behaves from studying \eqref{eq1}! It increases beyond bounds, so our left hand side must to do so as well, and thus can't converge! We get a second partial solution:

$$\lim_{n\to\infty}\left(q+\frac1{n^2+1}\right)^{n^2} \text{ does not exist, if } q < -1. $$

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The remaining case is $q=-1$, which means

$$\lim_{n\to\infty}\left(-1+\frac1{n^2+1}\right)^{n^2} = \lim_{n\to\infty}(-1)^{n^2}\left(1-\frac1{n^2+1}\right)^{n^2} \tag{4}\label{eq4}$$

The first factor is alternating betwee $+1$ and $-1$ the second factor should make you feel you've seen it somewhere, it is very similiar to the following limit, which you should know:

$$\lim_{n\to\infty}\left(1-\frac1n\right)^n=e^{-1}.$$

From this we can just take the subsequence where we only consider the indices of the form $n^2+1$:

$$\lim_{n\to\infty}\left(1-\frac1{n^2+1}\right)^{n^2+1}=e^{-1}.$$

Now the only difference to what we need is the exponent, but that can be handled easily by the normal rules for limits of quotients:

$$\lim_{n\to\infty}\left(1-\frac1{n^2+1}\right)^{n^2}=\lim_{n\to\infty}\frac{\left(1-\frac1{n^2+1}\right)^{n^2+1}}{\left(1-\frac1{n^2+1}\right)}=\frac{\lim_{n\to\infty}\left(1-\frac1{n^2+1}\right)^{n^2+1}}{\lim_{n\to\infty}\left(1-\frac1{n^2+1}\right)}=\frac{e^{-1}}1=e^{-1}.$$

To get back to \eqref{eq4}, we now have to consider the limit of a product, whe the first factor alternates between $+1$ and $-1$, and the second factor tends to $e^{-1}$. Since $e^{-1} \neq 0$, this can't converge, as all the odd sequence members converge to $-e^{-1}$ and all the even sequence members converge to $e^{-1}$. So we find that for $q=-1$, the sequence does not converge.

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Summing up, the sequence converges (to $0$) exactly for all $-1 < q \le 0$, which is equivalent to $0 < \sin(x) \le 1$ which is equivalent to $2k\pi < x <(2k+1)\pi$ for some $k \in \mathbb Z$.