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# ISI MStat 2018 PSA Problem 7 | Continuous Function

This is a problem from ISI MStat 2018 PSA Problem 7 based on Continuity.

## Continuous Function - ISI MStat Year 2018 PSA Question 7

Let $$f$$ be a function defined from $$(0, \infty)$$ to $$\mathbb{R}$$ such that
$$\lim _{x \rightarrow \infty} f(x)=1$$ and $$f(x+1)=f(x)$$ for all x
Then $$f$$ is

• (A) continuous and bounded.
• (B) continuous but not necessarily bounded.
• (C) bounded but not necessarily continuous.
• (D) neither necessarily continuous nor necessarily bounded.

### Key Concepts

Epsilon-Delta definition of limit

Continuity

Bounded function

ISI MStat 2018 PSA Problem 7

Introduction to Real Analysis by Bertle Sherbert

## Try with Hints

Try to use the epsilon-delta definition of limit and the property that $$f(x+1)=f(x)$$ for all x.

$$\lim_{x \to \infty} f(x) = 1$$ $$\Rightarrow \exists M > 0$$ such that $$x > M \Rightarrow$$ $$|f(x) - 1| < \epsilon$$. Now $$f(x) = f(x+1)$$ for all $$x \in \mathbb{R} \Rightarrow f(x) = f(x+n)$$ for every $$n \in \mathbb{Z}$$.

Let's try to use this .

Given any $$y \in \mathbb{R}$$ we can select a suitable $$n \in \mathbb{Z}$$ such that $$y+n > M$$. Then $$|f(y+n) - 1| < \epsilon$$. But $$f(y+n) = f(y)$$. Hence , $$|f(y) - 1| < \epsilon$$. Hence , for all $$y \in \mathbb{R},$$ we have $$|f(y) - 1| < \epsilon$$. Since , $$\epsilon > 0$$ is arbitrary , we must have $$f(y) = 1$$ for all $$y \in \mathbb{R}.$$

So $$f$$ is continuous and bounded

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