This is a problem from ISI MStat 2018 PSA Problem 7 based on Continuity.
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
Epsilon-Delta definition of limit
Continuity
Bounded function
But try the problem first...
Answer: is (A)
ISI MStat 2018 PSA Problem 7
Introduction to Real Analysis by Bertle Sherbert
First hint
Try to use the epsilon-delta definition of limit and the property that \( f(x+1)=f(x)\) for all x.
Second Hint
\( \lim_{x \to \infty} f(x) = 1 \) \( \Rightarrow \exists M > 0\) 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 .
Final Step
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
This is a problem from ISI MStat 2018 PSA Problem 7 based on Continuity.
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
Epsilon-Delta definition of limit
Continuity
Bounded function
But try the problem first...
Answer: is (A)
ISI MStat 2018 PSA Problem 7
Introduction to Real Analysis by Bertle Sherbert
First hint
Try to use the epsilon-delta definition of limit and the property that \( f(x+1)=f(x)\) for all x.
Second Hint
\( \lim_{x \to \infty} f(x) = 1 \) \( \Rightarrow \exists M > 0\) 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 .
Final Step
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
