If $f(x)=\lim_{t\to\infty}{\frac{(1+\sin{\pi x})^t-1}{(1+\sin{\pi x})^t+1}}$, then range of $f(x)$ is?

Question

If $$f(x)=\lim_{t\to\infty}{\frac{(1+\sin{\pi x})^t-1}{(1+\sin{\pi x})^t+1}}$$

Then range of $f(x)$ is?

My Attempt:

I was able to conclude that when,

$$\sin{\pi x}\to0^+, f(x)\to1$$

$$\sin{\pi x}\to0^-, f(x)\to-1$$

But the answer is $\{-1,0,1\}$

When will $f(x)$ assume the value $0$?

user · Accepted Answer · 2018-08-20T16:30:38.330

2

We have that

$$f(0)=\lim_{t\to\infty}{\frac{(1+\sin{0})^t-1}{(1+\sin{0})^t+1}}=\frac{1-1}{2}=0$$

edited Aug 20 '18 at 16:30

answered Aug 20 '18 at 16:29

user

154,566

Isn't $1^{\infty}$ an indeterminant form? – prog_SAHIL Aug 20 '18 at 16:30
@prog_SAHIL We have $\lim_{t \to \infty} 1^t=1$ since $1^a=1 \quad \forall a\in R$ – user Aug 20 '18 at 16:31
2

And $f(1/4)=1.$ And $f(-1/4)=-1.$ – mfl Aug 20 '18 at 16:33
Can we continue this in a chat? – prog_SAHIL Aug 20 '18 at 16:34
3

@prog_SAHIL We refer to indeterminate forms $1^{\infty}$ in cases like $(1+1/n)^n$ but here the base is identically $1$. – user Aug 20 '18 at 16:34
@prog_SAHIL Yes exactly! It is not an indeterminate form in that case. – user Aug 20 '18 at 16:36
Thanks for the help. – prog_SAHIL Aug 20 '18 at 16:39
@prog_SAHIL You are welcome! Bye – user Aug 20 '18 at 16:39

score 1 · Answer 2 · answered Aug 20 '18 at 16:41

We have three possibilities: $\sin \pi x=0, \sin \pi x<0 $ and $\sin \pi x>0.$ Since we have that

$$\lim_{t\to \infty} \dfrac{(1+a)^t-1}{(1+a)^t+1}$$ is $0$ if $a=0$, $1$ if $a>0$ and $-1$ if $-1<a<0$ we can conclude that the range of $f$ is $\{-1,0,1\}.$

So $f$ takes the value $0$ if $\sin \pi x=0.$ That is $$x\in \{n\pi |n\in\mathbb{Z}\}. $$

Anastassis Kapetanakis · Answer 3 · 2018-08-20T18:06:39.473

1

Another way to solve the problem is the following!

$$\lim_{t\to \infty}\frac{(1+sinπx)^t-1}{(1+sinπx)^t+1}=1-\lim_{t\to\infty}\frac{2}{(1+sinπx)^t+1}=k$$

When $sinπx=-1$: $k=1-2=-1$

When $sinπx=0$: $k=1-1=0$

When $sinπx<0$: $k=1-2=-1$

When $sinπx>0$: $k = 1$

Hence the range of the function is ${-1,0,1}$

As you see the above function is equal to zero for every $x$ that $sinπx=0$

edited Aug 20 '18 at 18:06

answered Aug 20 '18 at 16:48

Anastassis Kapetanakis

1,020

It is not true. Note that $f(-\pi/4)=-1.$ It depends when $\sin \pi x=0$ and in its sign if it isn't $0.$ – mfl Aug 20 '18 at 17:20
@mfl Yes that's what I am actually saying in my answer!!! – Anastassis Kapetanakis Aug 20 '18 at 18:03
No. $\sin (-\pi/4)=-\sqrt{2}/2$ is inside "For any other". And according to your answer the limit must be $1.$ But it is $-1$. – mfl Aug 20 '18 at 18:06
@mfl Yeah I noticed that mistake right now and corrected that!! Thanks for pointing that out!! – Anastassis Kapetanakis Aug 20 '18 at 18:07

If $f(x)=\lim_{t\to\infty}{\frac{(1+\sin{\pi x})^t-1}{(1+\sin{\pi x})^t+1}}$, then range of $f(x)$ is?

3 Answers3