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- anonymous

did i do this correctly? lim sqrt(x^2+1)
x->infinity----------
x+1
it's hard to type this so i'll explain. I divided the sqrt x^2 +1/ x^2. Then i put x+1/x because the highest power of x in the denominator is x^1. My answer is 1...would it also be 1 if x-> -infinity?

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- anonymous

- schrodinger

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- anonymous

So, it seems you mean \[ \Large{\lim_{x \rightarrow \infty} \dfrac {\sqrt {(x^2 + 1)}}{x + 1}}. \]I would use l'Hopital's rule, since we have infty/infty (an indeterminate form). We have that the derivative of the numerator is \[ \dfrac {2x}{\sqrt {(x^2 + 1)}} \] and the derivative of the denominator is 1. So, the limit becomes: \[ \Large{\lim_{x \rightarrow \infty} \dfrac {2x}{\sqrt{(x^2 + 1)}}}. \]Also, note that the original limit is the same as if the limit was of \[ \dfrac {\sqrt {(x^2 + 1)}}{x}, \] since the limit is taken as x goes to infty. Since the reciprocal of this limit is equal to 1/2 this limit, we can say that \[ L = \dfrac {2}{L}, \]if L is the original limit. Thus, the limit is either sqrt(2) or -sqrt(2). Since it seems that it would be positive for +infty and negative for -infty, the answer to your first question would be \[ \Large{\boxed{\sqrt{2}}} \]and "No. -sqrt(2)" for your second question.
Of course, I'm not an expert yet, so I'll recommend that either @CliffSedge or @satellite73 should check my work.

- anonymous

I'd have to get clarification from @shushu55 before I offered any advice.

- anonymous

@shushu55;whats the question???

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