anonymous
  • anonymous
find the real constants of a and b such that the limit of √a^x2 + bx −√x^2 + x + 1 goes to 1 when x approaches positive infinity
Calculus1
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Please find the attached file for actual question
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anonymous
  • anonymous
\[\lim_{x\to\infty}\left(\sqrt{ax^2+bx}-\sqrt{x^2+x+1}\right)\] Multiply by the conjugate: \[\lim_{x\to\infty}\left(\sqrt{ax^2+bx}-\sqrt{x^2+x+1}\right)\cdot\frac{\sqrt{ax^2+bx}+\sqrt{x^2+x+1}}{\sqrt{ax^2+bx}+\sqrt{x^2+x+1}}\] \[\lim_{x\to\infty}\frac{ax^2+bx-\left(x^2+x+1\right)}{\sqrt{ax^2+bx}+\sqrt{x^2+x+1}}\] \[\lim_{x\to\infty}\frac{(a-1)x^2+(b-1)x-1}{\sqrt{x^2}\left(\sqrt{a+\frac{b}{x}}+\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}\right)}\] \[\lim_{x\to\infty}\frac{(a-1)x^2+(b-1)x-1}{|x|\left(\sqrt{a+\frac{b}{x}}+\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}\right)}\] Since \(x\to\infty\), you have \(x>0\), so \(|x|=x\): \[\lim_{x\to\infty}\frac{(a-1)x^2+(b-1)x-1}{x\left(\sqrt{a+\frac{b}{x}}+\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}\right)}\] \[\lim_{x\to\infty}\frac{(a-1)x+(b-1)-\frac{1}{x}}{\sqrt{a+\frac{b}{x}}+\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}}\] Now, as \(x\to\infty\), all the terms with \(x\) in the denominator will approach 0, leaving you with \[\lim_{x\to\infty}\frac{(a-1)x+b-1}{\sqrt{a}+1}\]
anonymous
  • anonymous
The limit can only be finite if the \(x\) term disappears, so you have \(a=\cdots\). Should be easy to solve for \(b\) from there.

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anonymous
  • anonymous
the limit is 1/ So here a = 1, b = 3, rihgt?
anonymous
  • anonymous
Yep, that's what I'm getting.
anonymous
  • anonymous
thanks! another question
anonymous
  • anonymous
need to prove it using precise definiton
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