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Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a continuous function such that $\lim _{n\rightarrow \infty} f’’(x)$ exists for every $x \in \mathbb{R}$, where $f’’(x) = f  \circ f^{n-1}(x)$ for $n \geq 2$. Define 

$$S=\left\{\lim _{n \rightarrow \infty} f’’(x): x \in \mathbb{R}\right\}  \text{ and } T=\left\{x  \in \mathbb{R}:f(x)=x\right\}$$

Then which of the following is necessarily true?

  1. $S \subset T$
  2. $T \subset S$
  3. $S = T$
  4. None of the above
in Calculus by
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2 Answers

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$h(x)  = \lim_{n \rightarrow \infty} f^n(x) = f(\lim_{n \rightarrow \infty}f^{n-1}(x))$

We can perform the last step as the function is continuous for every x. Continuing on,

$h(x) = f(h(x))$ $ \forall x \in R$

because $n\rightarrow \infty\implies n-1 \rightarrow \infty$

It means $S = \{f(x) = x : \forall x \in Range(h(x))\}$

The function defined by two sets in same but $S$ is defined over some elements of real numbers whereas $T$ is defined for all real numbers.

$\implies S \subset T$

$A$ is correct answer.
How do u know $T$ mapped on all real numbers. It also can be mapped on some real numbers
Given in question. f is defined on all real numbers because domain is R . and most importantly in the definition of T,  $ x\in R$ which means no restriction. And this relation holds necessarily. But it could be the case that both S and T are equal if $Range(h(x)) = R$ , not otherwise.
0 votes
f^{n}(x)=f \circ f^{n-1}(x)

$taking$  n = 2 ,

putting the value of $f(x)=x$

S={f(x), where  $x \in \mathbb{R}$}

So T is a subset of S

*Correct me if I am wrong
edited by

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