Limits Problem: lim (x^2 - 4 - 3 x)^1/2 - x as x -> +inft

[Aduial.Elen]

I'm doing a calculus assignment and I'm stumped by the last question- how frustrating!


lim (x^2 - 4 - 3 x)^1/2 - x
x -> +oo

x^2 indicates x squared
^1/2 indicated that the entire bracketed section has an exponent of one half
x -> +oo incidcates the x values go to infinity

I just don't know what that - x is doing on the back, or what to do with it. Any help would be VERY much appreciated!

 

[pka]

\(\displaystyle \L
\begin{array}{rcl}
\sqrt {x^2 - 3x - 4} - x & = & \left( {\sqrt {x^2 - 3x - 4} - x} \right)\left( {\frac{{\sqrt {x^2 - 3x - 4} + x}}{{\sqrt {x^2 - 3x - 4} + x}}} \right) \\
& = & \frac{{ - 3x - 4}}{{\sqrt {x^2 - 3x - 4} + x}} \\
& = & \frac{{ - 3 - \frac{4}{x}}}{{\sqrt {1 - \frac{3}{x} - \frac{4}{{x^2 }}} + 1}} \\
\end{array}\)

 

[Aduial.Elen]

I got to your third step, so not too bad! So the answer's -3/2 I guess. I was close before but not quite correct.

Wait- I don't quite understand your last step. You divided the top values by x but the bottom values by x^2. How are you allowed to do that??

 

[galactus]

He divided by \(\displaystyle \sqrt{x^{2}}\)

 

[Aduial.Elen]

Ok, thank you- it makes sense now! =]