Limits help!!

[mathkid]

Limits help!! *urgent*

I do not for the life of me understand how to do this hw problem.... we were told the answer is 1/2 but I keep thinking it should be zero

limit (as x goes to infinity) of x^2 - sqrt( x^4 - x^2 + 2)

help?

 

[pka]

the answer is 1/2 but I keep thinking it should be zero
limit (as x goes to infinity) of x^2 - sqrt( x^4 - x^2 + 2)

\(\displaystyle \begin{gathered} x^2 - \sqrt {x^4 - x^2 + 2} \\
= \dfrac{{x^2 - 2}}{{x^2 + \sqrt {x^4 - x^2 + 2} }} \\
= \frac{{1 - \frac{2}{{x^2 }}}}{{1 + \sqrt {1 - \frac{1}{{x^2 }} + \frac{2}{{x^4 }}} }} \\ \end{gathered} \)

 

[mathkid]

\(\displaystyle \begin{gathered} x^2 - \sqrt {x^4 - x^2 + 2} \\
= \dfrac{{x^2 - 2}}{{x^2 + \sqrt {x^4 - x^2 + 2} }} \\
= \frac{{1 - \frac{2}{{x^2 }}}}{{1 + \sqrt {1 - \frac{1}{{x^2 }} + \frac{2}{{x^4 }}} }} \\ \end{gathered} \)

Sorry, this is probably just my lack of algebra skills talking... but how'd you get to the first step after the original equation?
thanks

 

[mathkid]

actually never mind. You multiplied top and bottom by the conjugate I believe. Thank you so much!

 

[soroban]

Hello, mathkid!

This one requires some Olympic-level gymnastics . . .

\(\displaystyle \displaystyle \lim_{x\to\infty}\,\left(x^2 - \sqrt{x^4-x^2+2}\right)\)

Mutiply "top and bottom" by the conjugate:

\(\displaystyle \dfrac{x^2-\sqrt{x^4-x^2+2}}{1}\cdot\dfrac{x^2+\sqrt{x^4-x^2+2}}{x^2+\sqrt{x^4-x^2+2}}\)

. . . . \(\displaystyle =\;\dfrac{x^4 - (x^4-x^2+2)}{x^2+\sqrt{x^4-x^2+2}} \;=\; \dfrac{x^2-2}{x^2+\sqrt{x^4-x^2+2}} \)


Divide numerator and denominator by \(\displaystyle x^2\!:\)

\(\displaystyle \dfrac{\frac{x^2}{x^2} - \frac{2}{x^2}}{\frac{x^2}{x^2} + \frac{\sqrt{x^4-x^2+2}}{x^2}} \;=\;\dfrac{1 - \frac{2}{x^2}}{1 + \sqrt{\frac{x^4-x^2+2}{x^4}}} \)

. . . . \(\displaystyle =\;\dfrac{1 - \frac{2}{x^2}}{1 + \sqrt{\frac{x^4}{x^4} - \frac{x^2}{x^4} + \frac{2}{x^4}}} \;=\;\dfrac{1-\frac{2}{x^2}}{1 + \sqrt{1 - \frac{1}{x^2} +\frac{2}{x^4}}} \)


Therefore: .\(\displaystyle \displaystyle\lim_{x\to\infty}\dfrac{1-\frac{2}{x^2}}{1 + \sqrt{1 - \frac{1}{x^2} +\frac{2}{x^4}}} \;=\; \frac{1-0}{1 + \sqrt{1+0+0}} \;=\;\frac{1}{1+1} \;=\;\frac{1}{2}\)

 

[lookagain]

Hello, mathkid!

This one requires ? some Olympic-level gymnastics . . .

I will be using the Binomial Theorem.

There are to be limit notations placed at the front of all of the expressions
in the following steps, except for the final line.



\(\displaystyle x^2 \ - \ [x^4 + (-x^2 + 2)]^{1/2} \ = \)


\(\displaystyle x^2 \ - \bigg[\ (x^4)^{1/2} \ + \ \dfrac{1}{2}\bigg(\dfrac{1}{1!}\bigg)[(x^4)^{-1/2}](-x^2 + 2)^1 \ + \ \dfrac{1}{2}\bigg(\dfrac{-1}{2}\bigg)\bigg(\dfrac{1}{2!}\bigg)[(x^4)^{-3/2}](-x^2 + 2)^2 \ + \ ... \bigg] \ = \)


\(\displaystyle x^2 \ - \ \bigg[x^2 \ + \ \dfrac{1}{2x^2}\bigg(\dfrac{-x^2 + 2}{1}\bigg) \ - \ \dfrac{1}{8}\bigg(\dfrac{1}{x^6}\bigg)\dfrac{(-x^2 + 2)^2}{1} \ + \ ...\ \bigg]= \)


\(\displaystyle x^2 \ - \bigg[x^2 \ - \dfrac{1}{2} \ + \ \dfrac{1}{x^2} \ + \ (the \ \ rest \ \ of \ \ the \ \ terms \ \ have \ \ degree \ \ less \ \ than \ \ or \ \ equal \ \ to \ \ negative \ \ two)\bigg] \ = \)


\(\displaystyle \dfrac{1}{2} \ - \ \dfrac{1}{x^2} \ - \ (the \ \ rest \ \ of \ \ the \ \ terms \ \ have \ \ degree \ \ less \ \ than \ \ or \ \ equal \ \ to \ \ negative \ \ two) \ \)


\(\displaystyle Then \ \ the \ \ limit, \ \ as \ \ x \ \ approaches \ \ \infty, \ \ of \ \ the \ \ above \ \ expression \ \ equals \ \ \dfrac{1}{2}.\)