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Resolving $ \lim_{x\to 0} \frac{x\ln x}{(x+1)^2}\; $ using L'Hospital's rule we obtain that:

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

but I'm not sure if that's enough for the function to be continuous and when I graph it, it seems to be.

Harish Chandra Rajpoot
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jackes gamero
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    Using l'Hopital on what? $\lim_{x\to 0}(x+1)^2$ is neither $0$ nor $\infty$. –  May 04 '20 at 17:11
  • $\lim_{x\to 0} \frac{xln x}{(x+1)^2} $ = $ \frac{\lim_{x\to 0} xln x}{\lim_{x\to 0} (x+1)^2} $ = $ \frac{\lim_{x\to 0} ln x/x}{\lim_{x\to 0} (x+1)^2} $ = $ \lim_{x\to 0} \frac{ln x}{x} $ – jackes gamero May 04 '20 at 17:13
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    $f$ isn't defined at $0,$ so I would say no. – zhw. May 04 '20 at 17:14
  • To be clear, OP, the fact that $f$ is not defined means that technically your question doesn't make sense, not that the answer to your question is no. If we make the obvious definition $f(0) = 0$, then $f$ is continuous in this interval, since as your limit shows, it is continuous at $0$, and by basic composition theorems about continuity, and the fact that polynomials and $\ln$ are continuous, it is continuous everywhere else. – Izaak van Dongen May 04 '20 at 17:26

5 Answers5

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The function is not defined at $x = 0$ hence can not be continuous there. This means the limit you had is not enough .

Bernard
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DeepSea
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It's continuous on $(0,1]$ for certain, but it's not defined at $x=0$. But if we can continuously extend it to $x=0^+$, then it'll be continuous on $[0,1]$. I'd avoid using LHR since a priori we don't know if it's continuous (much less differentiable). We have $$ \lim_{x\to 0^+} \frac{x\ln(x)}{(x+1)^2} $$It suffices to show $ \lim\limits_{x\to 0^+}{x\ln(x)}$ exists. Make the substitution $x=e^{-u}$: $$ \lim\limits_{x\to 0^+}{x\ln(x)} = \lim\limits_{u\to\infty}{e^{-u}\ln(e^{-u})}=-\lim\limits_{u\to\infty}{ue^{-u}} $$Now we make two observations:

  • For $u>0,$ $u e^{-u}>0$
  • For $u>1,$ $e^u > u^2$ (this follows by convexity)

Then by the Squeeze Theorem we have $$ -\lim\limits_{u\to\infty}{u\cdot u^{-2}} \leq -\lim\limits_{u\to\infty}{ue^{-u}} \leq 0 $$ $$ 0 \leq -\lim\limits_{u\to\infty}{ue^{-u}} \leq 0 $$So the limit exists and is $0$. Then if we define $f(0)=0$, we get that $f$ is continuous on $[0,1]$.

Integrand
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What is true is that $f$, which is defined only on $(0,+\infty)$ can be continuously extended to $[0,+\infty)$ because we have the well-known high-school limit $$\lim_{x\to 0^+}x\ln x=0,$$ derived by substitution from $\lim_{x\to+\infty}\smash{\dfrac{\ln x}x=0}$.

Therefore, setting $f(0)=0$, we obtain a continuous function.

Bernard
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I could be wrong, but I suspect the question you were trying to ask is:

"What is $ \lim_{x\to 0^+} \frac{x\ln x}{(x+1)^2}\ $?"

If so, see:

How do I prove that $\lim_{x→0} x⋅\ln x=0$

Adam Rubinson
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In general, for a function $f(x)$ to be continuous at any point $x=a$, we must have $$\lim_{x\to a^{-}}f(x)=\lim_{x\to a^{+}}f(x)=f(a)$$

But for given function $f(x)=\frac{x\ln x}{(x+1)^2}$, we have $$\lim_{x\to 0^{-}}\frac{x\ln x}{(x+1)^2}=\lim_{x\to 0^{+}}\frac{x\ln x}{(x+1)^2}=0\ne f(0)$$ Since given function is not defined at $x=0$ hence is not continuous at $x=0$

Harish Chandra Rajpoot
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