Question

Find
lim_{x rightarrow infty} (sqrt{x^2+5x+1}-x) [hint: rationalize.]

Answer 1

\[\lim_{x \rightarrow \infty}(\sqrt{x^2+5x+1}-x)\]

Answer 2

hint:
we can write:
\[\large \sqrt {{x^2} + 5x + 1} - x = \frac{{{x^2} + 5x + 1 - {x^2}}}{{\sqrt {{x^2} + 5x + 1} + x}}\]

Answer 3

more explanation:
\[\large \begin{gathered}
\sqrt {{x^2} + 5x + 1} - x = \frac{{\left( {\sqrt {{x^2} + 5x + 1} - x} \right)\left( {\sqrt {{x^2} + 5x + 1} + x} \right)}}{{\sqrt {{x^2} + 5x + 1} + x}} = \hfill \\
= \frac{{{x^2} + 5x + 1 - {x^2}}}{{\sqrt {{x^2} + 5x + 1} + x}} \hfill \\
\end{gathered} \]

Answer 4

Yes, I got that part

Answer 5

ok! Then, we have to factor out x at numerator and at denominator, so we get:
\[\frac{{5x + 1}}{{\sqrt {{x^2} + 5x + 1} + x}} = \frac{{x\left( {5 + 1/x} \right)}}{{x\left( {\sqrt {1 + 5/x + 1/{x^2}} + 1} \right)}} = \frac{{5 + 1/x}}{{\sqrt {1 + 5/x + 1/{x^2}} + 1}}\]

Answer 6

that expression is correct if
\[\large x \to + \infty \]

Answer 7

oh okay

Answer 8

whereas if \[x \to - \infty \]
we have:
\[\frac{{5x + 1}}{{\sqrt {{x^2} + 5x + 1} + x}} = \frac{{x\left( {5 + 1/x} \right)}}{{ - x\left( {\sqrt {1 + 5/x + 1/{x^2}} + 1} \right)}} = - \frac{{5 + 1/x}}{{\sqrt {1 + 5/x + 1/{x^2}} + 1}}\]

Answer 9

Just wondering where did you get the 1/x^2 in the denominator?

Answer 10

since we have the subsequent steps:
\[\large \sqrt {{x^2} + 5x + 1} + x = \sqrt {{x^2}\left( {1 + \frac{5}{x} + \frac{1}{{{x^2}}}} \right)} + x = \left| x \right|\sqrt {1 + \frac{5}{x} + \frac{1}{{{x^2}}}} + x\]

Answer 11

\[\large \begin{gathered}
\sqrt {{x^2} + 5x + 1} + x = \sqrt {{x^2}\left( {1 + \frac{5}{x} + \frac{1}{{{x^2}}}} \right)} + x = \hfill \\
= \left| x \right|\sqrt {1 + \frac{5}{x} + \frac{1}{{{x^2}}}} + x \hfill \\
\end{gathered} \]

Answer 12

please, I have made an error, when \[x \to - \infty \]
the original expression, can be rewritten as follows:
\[\frac{{5x + 1}}{{\sqrt {{x^2} + 5x + 1} + x}} = \frac{{x\left( {5 + 1/x} \right)}}{{x\left( { - \sqrt {1 + 5/x + 1/{x^2}} + 1} \right)}} = \frac{{5 + 1/x}}{{ - \sqrt {1 + 5/x + 1/{x^2}} + 1}}\]

Answer 13

Oh I see... Then when an sub in the infinity and follow limit law and get rid of the 1/x, 5/x and 1/x^2 beacuse they will equal to zero, right? So we get 5/sqrt 2? @Michele_Laino

Answer 14

I mean 5/2

Answer 15

yes!

Answer 16

YAY! Thank you so much! :)

Answer 17

Thank you! :)