0

let $f: (-a,a) \setminus \{0\} \longrightarrow \mathbb{R}$. show that $\lim_{x\to 0} f(x) = L$ iff $\lim_{x\to 0} f(\sin x) = L$.

Could anyone give me a hint please?

Wisław
  • 463
Intuition
  • 3,269
  • 1
    Assuming that $f$ is contiuous, what happens when you take limit inside of the function? That is, by continuity $\lim_{x\to 0} f(\sin x) = f(\lim_{x\to 0} \sin x)$. How are the limits $\lim_{x\to 0} \sin x$ and $\lim_{x\to 0} x$ related? – Wisław Oct 23 '18 at 09:33
  • I do not know how they are related @Wisław – Intuition Oct 23 '18 at 09:39
  • Well, $\lim_{x\to 0} \sin x = 0$. One motivation is to look at the Taylor series of $\sin x$ around $x = 0$. – Wisław Oct 23 '18 at 09:46
  • I want to prove it without Taylor series @Wisław – Intuition Oct 23 '18 at 09:51
  • See the theorem mentioned at the end of this answer https://math.stackexchange.com/a/1073047/72031 – Paramanand Singh Oct 23 '18 at 12:05
  • @Wisław: continuity of $f$ is not needed. The fact that $\sin x$ is invertible in neighborhood of $0$ is sufficient. See the answer linked in my previous comment. – Paramanand Singh Oct 23 '18 at 12:07

1 Answers1

1

Note that $\lim_{x \to 0} f(x) = L$ is equivalent to the statement : for every sequence in the domain $x_n \to 0$(where $\to$ means converging to as a sequence), we have $f(x_n) \to L$. Similarly, $\lim_{x \to 0} f(\sin x) = L$ is equivalent to the statement : for every $x_n \to 0$ we have $f(\sin x_n) \to L$.

What you should be knowing, is the fact that the sine function is continuous at zero : that is , if $x_n \to 0$ then $\sin x_n \to 0$. It follows that the inverse of the sine, defined on the range $[-1,1]$ of the sine function and going to the neighbourhood $[-0.5\pi,0.5\pi]$, is also continuous.

So one way is clear : suppose that $\lim_{x \to 0} f(x) = L$. Then, consider some $y_n \to 0$. We want to show that $f(sin y_n) \to 0$. But then, if $y_n \to 0$, we have $\sin y_n \to 0$, and therefore by the equivalent statement of $\lim_{x \to 0} f(x) = L$ which can be applied on any sequence converging to zero we get $f(\sin y_n) \to L$. Since this is true for all $y_n$ we conclude that $\lim_{x \to 0} f(\sin x) = L$.

The other way : Suppose that $\lim_{x \to 0} f(\sin x) = L$, and consider $y_n \to 0$. We want to show that $f(y_n) \to 0$. But how do we pull $y_n$ to create a sequence like $\sin x_n$?

The answer is to use the fact that $y_n \to 0$, which is inside $[-0.5\pi,0.5\pi]$. By definition of sequence convergence, $y_n$ lies eventually inside this interval. That is, there exists $n \in \mathbb N$ such that $y_n $ is in this interval for $n> N$. For these $y_n$, there exists $x_n$ such that $y_n = \sin x_n$, because $y_n$ lie in the range of the inverse sine function. But then, $y_n \to 0$ implies that $\sin x_n \to 0$, which by continuity of the inverse sine implies that $x_n \to 0$. Now, can you use the equivalence to finish?

Sarvesh Ravichandran Iyer
  • 76,105