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Calculate the Limit $f(x)$ of a sequence of functions $(f_x(x))_{n=1}^{\infty }$. We know $f_n(x)=\frac{x^{2n}}{1+x^{2n}}$.

My solution:

$$f(x)=\lim_{n\to\infty }\frac{x^{2n}}{1+x^{2n}}$$

for $x<1$ is $x^{2n}=0$ so

$$f(x)=\lim_{n\to\infty }\frac{x^{2n}}{1+x^{2n}}=0$$

for $x>1$ is $x^{2n}= \infty$ so

$$f(x)=\lim_{n\to\infty }\frac{x^{2n}}{1+x^{2n}}=\lim_{n\to\infty }\frac{1}{1+\frac{1}{x^{2n}}}=1$$

for $x=1$ I really don't know, help please. I tried to use L'hospital's rule (Derivative according to $n$) and I received something crazy

$$f(x)=\lim_{n\to\infty } \frac{0}{-2x^{-2n}ln(x)}$$

The problem is that for $x=1$ holds $x^{2n}=x^\infty=indeterminate$

Anakin
  • 223

2 Answers2

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For $x=1$, $f_n(x) = \frac{1}{2}\;\forall n$. If you want to calculate the pointwise convergence of $f_{n}(x)$ then you are done (at least for positive $x$, although negative x are trivial).

Uniform convergence may be a bit trickier however, but my guess is that if you do something like $f_{n}(1+1/(2n))$ or something similar you will see that you are always too far of 1 and therefore you don't get uniform convergence.

bernatguillen
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$$f_n(x)=\frac{x^{2n}}{1+x^{2n}}$$

$$x=0 $$ $$f_n(0)\rightarrow 0\quad\forall n$$

$$x=1$$

$$f_n(1)\rightarrow \frac{1}{2}\quad\forall n$$

$$x=-1$$

$$f_n(-1)\rightarrow \frac{1}{2}\quad\forall n$$

$$|x|> 1$$

$$lim_{n\rightarrow\infty}\frac{x^{2n}}{1+x^{2n}}=lim_{n\rightarrow\infty}\frac{1}{\frac{1}{x^(2n)}+1}=1$$

$$Thus,$$

$$f_n(x)\rightarrow f(x) =\begin{cases}0 &if& x = 0\\\frac{1}{2} & if&|x| = 1\\1&if&|x|>1\end{cases} $$

$$f_n\rightarrow f\neq 0\quad so\,not\,uniform\,convergent$$

chowching
  • 815