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# ISI MStat 2018 PSA Problem 7 | Continuous Function

This is a beautiful problem from ISI MSAT 2018 PSA problem 7 based on Continuous Funtion. We provide sequential hints so that you can try .

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