What is the value of $\lim_{x \to 1^+} \frac{\sqrt{x^2-1}+x-1}{ \sqrt{4x-4}+x^2-1}$? $$\lim_{x \to 1^+} \frac{\sqrt{x^2-1}+x-1}{ \sqrt{4x-4}+x^2-1} = ?$$
I have done these steps to find the answer:


*

*$x^2-1=(x+1)(x-1)$

*$\sqrt{4x-4}=2\sqrt{x-1}$

*$\displaystyle\lim_{x \to 1^+} \frac{\sqrt{x^2-1}+x-1}{ \sqrt{4x-4}+x^2-1} = \lim_{x \to 1^+} \frac{\sqrt{ (x+1)(x-1) }+x-1}{ 2\sqrt{x-1} + (x+1)(x-1) }$
So how do I remove what causes the hole function not to become $\frac{0}{0}$ and solve the limit?
 A: The expression is not of the form $0/0$ notice carefully the denominator goes to $0$ where the numerator is finite.
$$\lim_{x\to 1^+}\dfrac{\sqrt{x^2-1}+x+1}{\sqrt{4x-4}+x^2-1}\to \infty$$
This graph confirms the result: https://www.desmos.com/calculator/ejj9xiyjcf

After the OP's edit: $$\dfrac{\sqrt{x^2-1}+x-1}{2\sqrt{x-1}+x^2-1}=\dfrac{\sqrt{x-1}}{\sqrt{x-1}}\cdot\dfrac{\sqrt{x+1}+\sqrt{x-1}}{2+(x-1)^{3/2}(x+1)} \\ \implies \lim_{x\to 1^+}\dfrac{\sqrt{x+1}+\sqrt{x-1}}{2+(x-1)^{3/2}(x+1)}\to \dfrac{\sqrt{2}}{2}$$
A: I assume:
$$\lim_{x \to 1^+} \frac{\sqrt{x^2-1}+x\color{red}-1}{ \sqrt{4x-4}+x^2-1} = \lim_{x \to 1^+} \frac{\sqrt{x-1}\cdot \sqrt{x+1}+(\sqrt{x-1})^2}{ \sqrt{4x-4}+(\sqrt{x-1})^2(x+1)} =\\
\lim_{x \to 1^+} \frac{\sqrt{x-1}\cdot (\sqrt{x+1}+\sqrt{x-1})}{\sqrt{x-1}\cdot (2+(\sqrt{x-1})(x+1))} =\frac{\sqrt{2}}{2}.$$
A: Set $x-1=t^2$, which you can because the limit is for $x\to1^+$. Then the limit becomes
$$
\lim_{t\to0^+}\frac{t\sqrt{t^2+2}+t^2}{2t+t^2(t^2+2)}=
\lim_{t\to0^+}\frac{\sqrt{t^2+2}+t}{2+t(t^2+2)}
$$
