hpfan101
  • hpfan101
\[\lim_{x \rightarrow \infty}(\sqrt{x^2-ax}-\sqrt{x^2+bx}\]
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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jim_thompson5910
  • jim_thompson5910
Hint: Multiply top and bottom by the conjugate \[\large \lim_{x \rightarrow \infty}\left(\sqrt{x^2-ax}-\sqrt{x^2+bx}\right)\] \[\large \lim_{x \rightarrow \infty}\left[\left(\sqrt{x^2-ax}-\sqrt{x^2+bx}\right) \color{red}{\times \frac{\sqrt{x^2-ax}+\sqrt{x^2+bx}}{\sqrt{x^2-ax}+\sqrt{x^2+bx}}}\right]\]
anonymous
  • anonymous
wow am i wrong!
hpfan101
  • hpfan101
Alright so i multiplied by the conjugate and when I simplified I got \[\frac{ ax-bx }{ x \sqrt{ax}+x \sqrt{bx} }\]

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More answers

hpfan101
  • hpfan101
Apparently the answer is \[\frac{ 1 }{ 2 }(a-b)\]
jim_thompson5910
  • jim_thompson5910
you should get `-ax - bx` in the numerator when you expand things out
hpfan101
  • hpfan101
Though I don't know how to get to that
hpfan101
  • hpfan101
Oh ok, I'll change that
jim_thompson5910
  • jim_thompson5910
do you see how I got `-ax` ?
hpfan101
  • hpfan101
I'm assuming when you multiplied everything out by the conjugate. Though in my work, I don't see that. I must have made an error
hpfan101
  • hpfan101
Oh never mind I see what I did wrong
hpfan101
  • hpfan101
Can the equation be simplified further?
DanJS
  • DanJS
Calc 1 is 5% Calc shortcuts and 95% ALgebraaaa
jim_thompson5910
  • jim_thompson5910
I would factor out -x. Then multiply top and bottom by 1/x \[\Large \frac{-ax-bx}{\sqrt{x^2-ax}+\sqrt{x^2+bx}}\] \[\Large \frac{-x(a+b)}{\sqrt{x^2-ax}+\sqrt{x^2+bx}}\] \[\Large \color{red}{\frac{1/x}{1/x}\times}\frac{-x(a+b)}{\sqrt{x^2-ax}+\sqrt{x^2+bx}}\] \[\Large \frac{(1/x)*-x(a+b)}{(1/x)*\left(\sqrt{x^2-ax}+\sqrt{x^2+bx}\right)}\] \[\Large \frac{-1(a+b)}{(1/x)*\sqrt{x^2-ax}+(1/x)*\sqrt{x^2+bx}}\] Do you see how to finish up?
hpfan101
  • hpfan101
Yes I do. Thank you!
jim_thompson5910
  • jim_thompson5910
ok great
DanJS
  • DanJS
DanJS
  • DanJS
you doing the Delta-epsilon definition thing yet? I just reviewed that
hpfan101
  • hpfan101
Not yet. I just learned in class about the Intermediate Value Theorem. But thanks for the cheat sheet!
IrishBoy123
  • IrishBoy123
are you sure that is the right answer? \(\lim_{x \rightarrow \infty}(\sqrt{x^2-ax}-\sqrt{x^2+bx})\) \(=\lim_{x \rightarrow \infty}x(\sqrt{1-\frac{a}{x}}-\sqrt{1+\frac{b}{x}})\) using binomial expansion \(\sqrt{1+x} = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \frac{5}{128}x^4 + \frac{7}{256}x^5 - \cdots\)\ \(=lim_{x \rightarrow \infty}x( \{ 1 +\frac{1}{2}(-\frac{a}{x}) - \frac{1}{8}(-\frac{a}{x})^2 + \cdots \} - \{1 +\frac{1}{2}(\frac{b}{x}) - \frac{1}{8}(\frac{b}{x})^2\})\) \(= -\frac{1}{2}(a+b)\)
IrishBoy123
  • IrishBoy123
i mean this |dw:1443186003890:dw|
DanJS
  • DanJS
It has the negative in front, -(a+b)/2 From here \[\Large \frac{-x(a+b)}{\sqrt{x^2-ax}+\sqrt{x^2+bx}}\] take an x out of both roots in bottom and cancel it with the top to get \[\Large \frac{-(a+b)}{\sqrt{1-\frac{ a }{ x }}+\sqrt{1+\frac{ b }{ x }} }\] As x goes to infinity, those fractions go to zero a/x and b/x Left with the limit at infinity as \[\Large \frac{-(a+b)}{\sqrt{1}+\sqrt{1}} = \frac{-(a+b)}{2}\]
DanJS
  • DanJS
** \[\Large \lim_{x \rightarrow \infty } \frac{ 1 }{ x } = 0\]
IrishBoy123
  • IrishBoy123
yes @DanJS , just a typo i expect:-)
IrishBoy123
  • IrishBoy123
thx
DanJS
  • DanJS
what did you do, that looks like the taylor series expansion of that roots

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