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It seems like $\lim\limits_{x\rightarrow 0^{+}} \dfrac{x^{\min(a,b)}}{x^a+x^b}$ is $1$ if $a\ne b$, regardless of the values of $a$ and $b$, but is this true? What does this limit equal in general?

Thanks!

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4 Answers

up vote 4 down vote accepted

Wlog $a\le b$. Then your fraction is equal to $$\frac{1}{1+x^{(b-a)}}$$ If $b=a$ this is equal to $1/2$. If not, it tends to $1$ when $x$ tends to $0$.

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It never tends to $0$ as $x$ does. –  Brian M. Scott Dec 9 '11 at 18:58
    
@Brian: right you are. Corrected. Silly me. –  user20266 Dec 9 '11 at 19:02
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I should say that it is trivial. For $x\rightarrow 0$ you should keep the smaller power of $x$. That is it. But if $a\neq b$ then, for sure, $b>a$ and you should keep just terms containing $a$ and again you have 1.

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a! = b or a != b? –  Graphth Dec 9 '11 at 18:52
    
The question was edited. Indeed, it was not much meaningful. –  Jon Dec 9 '11 at 18:55
    
Gotcha!............ –  Graphth Dec 9 '11 at 19:09
    
Just to know. Who are you? A Tao's student or a common fool? –  Jon Dec 9 '11 at 20:37
    
I really have no idea what you're talking about. You wrote "a! = b", which means "a factorial = b". I was pretty sure you meant to write $a \neq b$, which is accomplished by the LaTeX "a \neq b" inside of dollar signs, so I asked. It looks like alex.jordan fixed this for you. And, this is your response to my asking a helpful question so that you could write what you actually meant to write? I don't even understand what a Tao's student is, and what is a "common" fool? Whatever you meant, I see I made a helpful suggestion and your response seems to be a jerk, unless I don't understand. –  Graphth Dec 10 '11 at 18:52
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Without loss of generality assume that $a\le b$. Then $$\lim_{x\to 0^+}\frac{x^a+x^b}{x^a}=1+\lim_{x\to 0^+}x^{b-a}=\begin{cases} 2,&b=a\\ 1,&b>a\;, \end{cases}$$

so $$\lim_{x\to 0^+}\frac{x^a}{x^a+x^b}=\begin{cases} \frac12,&b=a\\\\ 1,&b>a\;. \end{cases}$$

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HINT $\ $ It is easy to compute the limit of a quotient of two polynomials having equal order $\rm\:n\:$ because cancelling their common factor $\rm\:x^{\:n}\:$ leaves a determinate limit, namely

$$\rm\lim_{x\ \to\ 0}\ \frac{f_{\:n}\: x^{\:n} + f_{\:n+1}\ x^{\:n+1}+\ \cdots}{g_{\:n}\: x^n+g_{\:n+1}\: x^{\:n+1}\:+\ \cdots}\ =\ \lim_{x\ \to\ 0}\ \frac{f_{\:n} + f_{\:n+1}\: x\ +\ \cdots}{g_{\:n} +g_{\:n+1}\: x \:+\ \cdots}\ =\ \frac{f_n}{g_n}$$

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