<s>[INST] If $f(x)=\frac{\sqrt{x}-b}{x^2-1}$ and $\lim\limits_{x \to 1} f(x)=c$ then $b$ and $c$ are... If $f(x)=\frac{\sqrt{x}-b}{x^2-1}$ and $\lim\limits_{x \to 1} f(x)=c$ then $b$ and $c$ are...
I've been trying to find the limit as usual:
$\lim\limits_{x \to 1} \frac{\sqrt{x}-b}{x^2-1}=\lim\limits_{x \to 1} \frac{\sqrt{x}-b}{(x-1)(x+1)}$ and I still get zero in the denominator, so I don't know how to go from here...
 [/INST] $\lim\limits_{x \to 1} \frac{\sqrt{x}-b}{x^2-1}=\lim\limits_{x \to 1} \frac{\sqrt{x}-b}{(x-1)(x+1)}=\lim\limits_{x \to 1} \frac{\sqrt{x}-b}{(\sqrt{x}-1)(\sqrt{x}+1)(x+1)}$ so $b$ must be 1 and $c$ must be $1/4$
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