Rewriting Limit Problem with Conjugate - Explained | Get a Better Understanding

[CrossFit415]

x\stackrel{Lim}{\rightarrow}infinity (\sqrt{x^{2}+1})-1

Couldn't I just inset 0 for x and end up getting \sqrt{1}?

The book tells me to rewrite this problem by its conjugate.

My question is why? Thanks

 

[Mark44]

x\stackrel{Lim}{\rightarrow}infinity (\sqrt{x^{2}+1}-1

Couldn't I just inset 0 for x and end up getting \sqrt{1}?

The book tells me to rewrite this problem by the conjugate.

My question is why? Thanks

As you have written it,
\lim_{x \to \infty}\sqrt{x^2 + 1} - 1 = \infty
You seem to be a little confused in this problem. Why would you put in zero for x when the limit is as x goes to infinity?