Question

What is the limit as x approaches infinity, (sqrt(x^2+9x+1)-x)?

Answer 1

Now if the same rule applies I should end up with (sqrt(x^2(1+9/x+1/x^2))-x)

Answer 2

Multiply this expression by ( sqrt(x^2+9x+1) + x ) / ( sqrt(x^2+9x+1) + x )

And you'll see that this simplifies.

By the way, what is the answer intuitively of this limit?

Answer 3

i get (10x +1)/2x?

Answer 4

No.
(sqrt(x^2+9x+1)-x) . ( sqrt(x^2+9x+1) + x ) / ( sqrt(x^2+9x+1) + x )

= sqrt( x^2+9x+1)^2 - x^2 ) / ( sqrt(x^2+9x+1) + x ) , because the num was a diff
of squares

= ( x^2 + 9x + 1 - x^2 ) / ( sqrt(x^2+9x+1) + x )

= ( 9x + 1 ) / ( sqrt(x^2+9x+1) + x )

Now ... how do you handle this expression?

Answer 5

yea, i forgot that x cancels out so it would leave 9x+1/2x...therefore giving me 9/2...i believe

Answer 6

Yes, correct.

Answer 7

'good answer' if you don't mind.

Answer 8

Is this the question?
\[\lim_{x\rightarrow\infty}\sqrt{(x^2+9x+1)} -x\]

Answer 9

yes

Answer 10

\[\infty\]

Answer 11

no...it is 9/2

Answer 12

No Valpey. It's 9/2.

Answer 13

Follow the calculations above and you'll see why.

Answer 14

\[\lim_{x\rightarrow\infty}\sqrt{(x^2+9x+1)} - x\]
\[\lim_{x\rightarrow\infty}\frac{(\sqrt{(x^2+9x+1)} - x)(\sqrt{(x^2+9x+1)} + x )}{(\sqrt{(x^2+9x+1)} + x )}\]
\[\lim_{x\rightarrow\infty}\frac{(x^2+9x+1) - x^2}{(\sqrt{(x^2+9x+1)} + x )}\]
\[\lim_{x\rightarrow\infty}\frac{9x+1}{\sqrt{(x^2+9x+1)} + x}\]
\[\lim_{x\rightarrow\infty}\frac{9x+1}{x\sqrt{(1+9/x+1/x^2)} + x}\]
\[\lim_{x\rightarrow\infty}(\frac{9x}{x\sqrt{(1+9/x+1/x^2)} + x}+\frac{1}{x\sqrt{(1+9/x+1/x^2)} + x})\]
\[\lim_{x\rightarrow\infty}(\frac{9}{\sqrt{(1+9/x+1/x^2)} + 1}+\frac{1}{x\sqrt{(1+9/x+1/x^2)} + x})\]
\[(\frac{9}{\sqrt{(1)} + 1}+0) = \frac{9}{2}=4\frac{1}{2}\]