if $g(x)=\lim_{n→∞}\frac{\{x\}^{2n}-1}{\{x\}^{2n}+1}$ where $\{x\}=$fractional part of $x$, then find g(x).

Question

if $g(x)=\lim_{n→∞}\frac{\{x\}^{2n}-1}{\{x\}^{2n}+1}$

where $\{x\}=$fractional part of $x$,

then find $g(x)$.

My answer is coming out to be $g(x)=-1$. But I'm not sure since it's a constant function...Do i have to approximate the value a $g(x)$ as $x\to \infty$ and how to do so?

Answer 1

Let $x=y+t$, where $y\to \infty$ and $t\in[0,1]$.

Then

$$L=\lim_{y\to\infty}\frac{\{y+t\}^{2n}-1}{\{y+t\}^{2n}+1}=\frac{t^{2n}-1}{t^{2n}+1}$$

for any $t$ since $n$ is not specified.

But, as $n\to\infty$,

Then $$L=\lim_{n\to\infty}\frac{t^{2n}-1}{t^{2n}+1}=\frac{0-1}{0+1}=-1$$

Answer 2

I really think the answer depends on n more than x. As it involves frac(x) the value of that doesn't change even if x->infinity.

Before giving an answer take a look at this, maybe it gives you a hint. Sometimes graphing can actually help a lot.

https://www.desmos.com/calculator/ixvutwqnvz

As now it's edited in the question that it's n->infinity:

{x} will tend to zero as n->infinity due to the fact that {x}<1. And hence you did get the right answer.