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

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

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

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

My question is why? Thanks
 
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CrossFit415 said:
x[itex]\stackrel{Lim}{\rightarrow}[/itex]infinity ([itex]\sqrt{x^{2}+1}[/itex][itex]-1[/itex]

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

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

My question is why? Thanks
As you have written it,
[tex]\lim_{x \to \infty}\sqrt{x^2 + 1} - 1 = \infty[/tex]
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?