x to 1

There are definitely multiple ways to do this, but this works. For those who are familiar with " $\epsilon-\delta$ " (i.e. Weierstrass limit definition) limit proofs: $|x-1|<\delta\Rightarrow |x^2-1|=|x+1||x-1|<|x+1|\delta=|x-1+2|\delta<(|x-1|+2)\delta<(\delta+2)\delta$ . Since $\epsilon=\delta^2+2\delta$ is a one-to-one function for all positive $\epsilon$ when the domain of consideration is only the positive reals, then, for any given $\epsilon>0$ , we can simply choose $\delta$ to be the positive root of the equation $\delta^2+2\delta-\epsilon=0$ , which proves the limit is 1.

$\large \displaystyle \lim_{x\to 1} x^2 = 1^2 = \boxed{1}$

$\huge \displaystyle \lim_{x\to 1} x^2 = 1^2 = \boxed{\boxed{\boxed{\boxed{1}}}}$

it is determinate form we just put x=1 to get answer