Calculating $\lim\limits_{x\rightarrow 0^{+}} \frac{x^{\min(a,b)}}{x^a+x^b}$ 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!
 A: 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 $0$.
A: 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.
A: 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 $$\lim_{x\to 0^+}\frac{x^a}{x^a+x^b}=\begin{cases}
\frac12,&b=a\\\\
1,&b>a\;.
\end{cases}$$
A: 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
$$\rm\lim_{x\ \to\ 0}\ \frac{f_{\:n}\: x^{\:n} + f_{\:n+1}\ x^{\:n+1}+\ \cdots}{g_{\:n}\: x^n+g_{\:n+1}\: x^{\:n+1}\:+\ \cdots}\ =\ \lim_{x\ \to\ 0}\ \frac{f_{\:n} + f_{\:n+1}\: x\ +\ \cdots}{g_{\:n} +g_{\:n+1}\: x \:+\ \cdots}\ =\ \frac{f_n}{g_n}$$
