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hpfan101
 one year ago
\[\lim_{x \rightarrow \infty}(\sqrt{x^2ax}\sqrt{x^2+bx}\]
hpfan101
 one year ago
\[\lim_{x \rightarrow \infty}(\sqrt{x^2ax}\sqrt{x^2+bx}\]

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jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3Hint: Multiply top and bottom by the conjugate \[\large \lim_{x \rightarrow \infty}\left(\sqrt{x^2ax}\sqrt{x^2+bx}\right)\] \[\large \lim_{x \rightarrow \infty}\left[\left(\sqrt{x^2ax}\sqrt{x^2+bx}\right) \color{red}{\times \frac{\sqrt{x^2ax}+\sqrt{x^2+bx}}{\sqrt{x^2ax}+\sqrt{x^2+bx}}}\right]\]

hpfan101
 one year ago
Best ResponseYou've already chosen the best response.0Alright so i multiplied by the conjugate and when I simplified I got \[\frac{ axbx }{ x \sqrt{ax}+x \sqrt{bx} }\]

hpfan101
 one year ago
Best ResponseYou've already chosen the best response.0Apparently the answer is \[\frac{ 1 }{ 2 }(ab)\]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3you should get `ax  bx` in the numerator when you expand things out

hpfan101
 one year ago
Best ResponseYou've already chosen the best response.0Though I don't know how to get to that

hpfan101
 one year ago
Best ResponseYou've already chosen the best response.0Oh ok, I'll change that

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3do you see how I got `ax` ?

hpfan101
 one year ago
Best ResponseYou've already chosen the best response.0I'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
 one year ago
Best ResponseYou've already chosen the best response.0Oh never mind I see what I did wrong

hpfan101
 one year ago
Best ResponseYou've already chosen the best response.0Can the equation be simplified further?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0Calc 1 is 5% Calc shortcuts and 95% ALgebraaaa

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.3I would factor out x. Then multiply top and bottom by 1/x \[\Large \frac{axbx}{\sqrt{x^2ax}+\sqrt{x^2+bx}}\] \[\Large \frac{x(a+b)}{\sqrt{x^2ax}+\sqrt{x^2+bx}}\] \[\Large \color{red}{\frac{1/x}{1/x}\times}\frac{x(a+b)}{\sqrt{x^2ax}+\sqrt{x^2+bx}}\] \[\Large \frac{(1/x)*x(a+b)}{(1/x)*\left(\sqrt{x^2ax}+\sqrt{x^2+bx}\right)}\] \[\Large \frac{1(a+b)}{(1/x)*\sqrt{x^2ax}+(1/x)*\sqrt{x^2+bx}}\] Do you see how to finish up?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0you doing the Deltaepsilon definition thing yet? I just reviewed that

hpfan101
 one year ago
Best ResponseYou've already chosen the best response.0Not yet. I just learned in class about the Intermediate Value Theorem. But thanks for the cheat sheet!

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0are you sure that is the right answer? \(\lim_{x \rightarrow \infty}(\sqrt{x^2ax}\sqrt{x^2+bx})\) \(=\lim_{x \rightarrow \infty}x(\sqrt{1\frac{a}{x}}\sqrt{1+\frac{b}{x}})\) 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
 one year ago
Best ResponseYou've already chosen the best response.0i mean this dw:1443186003890:dw

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0It has the negative in front, (a+b)/2 From here \[\Large \frac{x(a+b)}{\sqrt{x^2ax}+\sqrt{x^2+bx}}\] 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
 one year ago
Best ResponseYou've already chosen the best response.0** \[\Large \lim_{x \rightarrow \infty } \frac{ 1 }{ x } = 0\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0yes @DanJS , just a typo i expect:)

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0what did you do, that looks like the taylor series expansion of that roots
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