0

If $f(x)=x\ln{x}$ for $x>0$ and $f(x)=0$ for $x=0$, then show that $f(x)$ is a real-valued, continuous function on $[0,\infty)$.

Is it enough too say the following: $\lim_{x\to0} f(x)=\lim_{x\to0} x\ln{x}=\lim_{x\to0} \frac{\ln{x}}{\frac{1}{x}}=\lim_{x\to0} \frac{\frac{1}{x}}{\frac{-1}{x^2}}=\lim_{x\to0} -x=0$

and $f(0)=0$,

and then

$\lim_{x\to{x_0}} f(x)=\lim_{x\to{x_0}} x\ln{x}=x_0ln{x_0}$

and $f(x_0)=x_0\ln{x_0}$

So $lim_{x\to{a}}=f(a)$ for $0\leq x$.

Thus, $f(x)$ is continuous for the desired interval.

sarahg
  • 1
  • 1
    Ok, what's the unique, single problematic point and what do you have to do in order to prove what you're being asked? – DonAntonio Mar 10 '14 at 12:02

1 Answers1

2

Notice $f$ is continuous everywhere in $(0,\infty)$. But, we still need to see what happens at $x = 0$. We claim that $f$ is indeed continuous at the origin. To see this. Let $\epsilon > 0$ be given. We want to find $\delta > 0$ such that if $|x| < \delta$, then $|f(x) - f(0)| < \epsilon$. But, with $|x| < 1$

$$ |f(x)| = |x \ln x | = x \ln x \leq x ( x - 1) < x -1 < \delta -1 $$

So, our choice $\delta = \min \{ 1, \epsilon + 1 \} $ would work.

In the first inequality, we have used:

$e^x \geq x +1 \implies e^{x-1} \geq x \implies x -1 \geq \ln x $.

  • Do you have to show that $f$ is continuous in $(0,\infty)$ as well or can you just say it? I mean, it obviously is continuous but how much do I have to say to show that it is? – sarahg Mar 10 '14 at 12:26
  • you can just say it since product of continuous functions is continuous. –  Mar 10 '14 at 12:27
  • ok, but then do you have to show that $\ln{x}$ and $x$ are continuous? What is just accepted as truth without justification? – sarahg Mar 10 '14 at 12:28
  • they are obviously continuous on $(0, \infty)$: PRoof: $\lim_{x \to a} x = a $ and $\lim_{x \to a } \ln x = \ln a $. –  Mar 10 '14 at 12:30
  • Ok, so one more question. Why can't I use the argument I wrote in my original post? What is wrong with it? – sarahg Mar 10 '14 at 12:32
  • Possibly for continuity at $0$ we should be explicit and showinstead that $\lim_{x\to 0^+} x\ln x=0$. – André Nicolas Mar 10 '14 at 12:32
  • @AndréNicolas Yes, thanks - that's what I meant. – sarahg Mar 10 '14 at 12:33