If $f(x)=\frac{\sqrt{x}-b}{x^2-1}$ and $\lim\limits_{x \to 1} f(x)=c$ then $b$ and $c$ are...

Question

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...

Answer 1

If $b, c\in \mathbb{R}$ is a condition we have that $$\lim_{x \to 1} x^2-1=0$$ Which is not so good, because it's in the denominator. The only way your fraction can have a limit is: $$\lim_{x \to 1} \sqrt{x}-b=0$$ So $b=1$ is the only possibility. Now you just need to compute the limit to get $c$: $$c=\lim_{x \to 1} \frac{\sqrt{x}-1}{x^2-1}$$

Answer 2

$\lim\limits_{x \to 1} \frac{\sqrt{x}-b}{x^2-1}=\lim\limits_{x \to 1} \frac{\sqrt{x}-b}{(x-1)(x+1)}=\lim\limits_{x \to 1} \frac{\sqrt{x}-b}{(\sqrt{x}-1)(\sqrt{x}+1)(x+1)}$ so $b$ must be 1 and $c$ must be $1/4$

Answer 3

Since $$f(x) =\frac{\sqrt{x} - b}{x^2-1}$$ it follows that $$b=\sqrt{x} - f(x) (x^2-1)$$ and taking limits as $x\to 1$ we get $$b=\sqrt{1}-c(1^2-1)=1$$ and now given $b=1$ it is easy to evaluate $$c=\lim_{x\to 1}f(x)=\lim_{x\to 1}\frac{\sqrt {x} - 1}{x^2-1}=\lim_{t\to 1}\frac{t - 1}{t^4-1}=\frac{1}{4}$$ In the above we have used the substitution $t=x^2$ and the standard limit $$\lim_{t\to a} \frac{t^n-a^n} {t-a} =na^{n-1}$$