Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Find the limits of the following functions. of the following functions. Find an interval on which convergence is uniform and another on which it is not.

$f_n(x) = (x/2)^n +(1/x)^n$


For $h_n(x) = (n+x)/(4n+x)$ show it converges uniformly on $[0,N]$ for any $N \lt \infty$ but not uniformly on $[0,\infty)$.

My text only has the theorem regarding uniform continuity so any help as to what algorithm is used for these type of questions would be greatly helpful

Thank you

share|cite|improve this question
Uniform continuity and uniform convergence are not the same thing. We say that a sequence of functions $f_n$ converge uniformly to a function $f$ on a set $E$ iff for all $\epsilon$ there exists an $n_0$ such that $$\sup_{x\in E}|f_n(x)-f(x)|<\epsilon$$ whenever $n\geq n_0$. Convergence then fails to be uniform if there is an $\epsilon$ such that for each $n$, there is an $x$ such that $|f_n(x)-f(x)|\geq\epsilon.$ Anyhow, you'll first want to figure out what the pointwise limit functions are--both how and where they are defined. Then you can start playing with uniform convergence. – Cameron Buie Feb 13 '13 at 4:11
up vote 1 down vote accepted

Hint: Assuming your $f_n:\mathbb{R}\rightarrow\mathbb{R}$, you see your $f_n(x)$'s are not defined at $x=0$, now regarding limit function that does not also exist while $x\in(0,1)$ because $\lim_{n\rightarrow}x^n\rightarrow 0$ while $x\in (0,1)$, but at $x=1,x=2$ limit exist and it is equal to $1$, clearly for $x>2$ the limit also does not exist.Now Analyze yourself.

$g_n(x)$'s are not defined at point $x=\frac{-2}{5n},n\in\mathbb{N}$, at $x=0$ limit does not exist, at $x=-n$ limit exists and equal to $0$,Now Analyze yourself.

$h_n(x)$'s are not defined at points $x=-4n$, in other case divide denominator and numerator by $n$ and see that limit function is $\frac{1}{4}$ as $n\rightarrow\infty$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.