<s>[INST] 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?
 [/INST] I assume:
$$\lim_{x \to 1^+} \frac{\sqrt{x^2-1}+x\color{red}-1}{ \sqrt{4x-4}+x^2-1} = \lim_{x \to 1^+} \frac{\sqrt{x-1}\cdot \sqrt{x+1}+(\sqrt{x-1})^2}{ \sqrt{4x-4}+(\sqrt{x-1})^2(x+1)} =\\
\lim_{x \to 1^+} \frac{\sqrt{x-1}\cdot (\sqrt{x+1}+\sqrt{x-1})}{\sqrt{x-1}\cdot (2+(\sqrt{x-1})(x+1))} =\frac{\sqrt{2}}{2}.$$
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