## shushu55 3 years ago 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?

1. Ahaanomegas

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.

2. CliffSedge

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

3. pasta

@shushu55;whats the question???