Answer
$$
\frac{x^{2}-1}{\sqrt{x+3}-2}
$$
Work Step by Step
A rational function $f$ is indeterminate at $x=1$ if the formula for $f(1)$ yields un undefined expression of the type $\dfrac{\infty}{\infty}$ or $\dfrac{0}{0}$.
We check $f(1)$ for each of the given functions.
$1)$ $f(x)=\dfrac{x^2+1}{x-1}$
$f(1)=\dfrac{1^2+1}{1-1}=\dfrac{2}{0}$ (not indeterminate)
$2)$ $f(x)=\dfrac{x^2-1}{x+2}$
$f(1)=\dfrac{1^2-1}{1+2}=\dfrac{0}{3}$ (not indeterminate)
$3)$ $f(x)=\dfrac{x^2-1}{\sqrt{x+3}-2}$
$f(1)=\dfrac{1^2-1}{\sqrt{1+3}-2}=\dfrac{0}{0}$ (indeterminate)
$4)$ $f(x)=\dfrac{x^2+1}{\sqrt{x+3}-2}$
$f(1)=\dfrac{1^2+1}{\sqrt{1+3}-2}=\dfrac{2}{0}$ (not indeterminate)
So the function is $\dfrac{x^2-1}{\sqrt{x+3}-2}$ is the indeterminate one.
