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limits

Calculating the limits of the above when x approaches infinity, why can't we just divide the function by the largest power of x and then say inside of the square root should be equal to "1" when x is at infinity level?

divided by the largest power of x

This seems so right to me but I know it is dead wrong. Could somebody tell me what I am missing?

user2694307
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  • The problem is you have neglected the $|x|$ on the outside of the radicals which blows up when $x$ tends to infinity – MPW Oct 03 '14 at 15:59
  • Yes, I totally got it now. I was thinking inside of the root as if x was at infinity, but at the same time and unfortunately, wasn't doing the same thing for the outside "abs(x)"s. Thanks for pointing it out. – user2694307 Oct 03 '14 at 16:06

1 Answers1

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Since $\lim_{x\to \infty}|x|\sqrt{1\pm\frac 2x}\to \infty$, we have $+\infty-\infty$. We cannot find the limit in this form.

Multiplying $$\sqrt{x^2+2x}-\sqrt{x^2-2x}$$ by $$\frac{\sqrt{x^2+2x}+\sqrt{x^2-2x}}{\sqrt{x^2+2x}+\sqrt{x^2-2x}}\ (=1)$$ gives you $$\lim_{x\to \infty}\frac{(\sqrt{x^2+2x}-\sqrt{x^2-2x})(\sqrt{x^2+2x}+\sqrt{x^2-2x})}{\sqrt{x^2+2x}+\sqrt{x^2-2x}}$$ $$=\lim_{x\to \infty}\frac{(x^2+2x)-(x^2-2x)}{\sqrt{x^2+2x}+\sqrt{x^2-2x}}=\lim_{x\to \infty}\frac{4x}{\sqrt{x^2+2x}+\sqrt{x^2-2x}}$$$$=\lim_{x\to\infty}\frac{\frac{4x}{x}}{\frac{\sqrt{x^2+2x}}{x}+\frac{\sqrt{x^2-2x}}{x}}=\lim_{x\to\infty}\frac{4}{\sqrt{\frac{x^2+2x}{x^2}}+\sqrt{\frac{x^2-2x}{x^2}}}$$$$=\lim_{x\to \infty}\frac{4}{\sqrt{1+\frac 2x}+\sqrt{1-\frac 2x}}=\frac{4}{\sqrt 1+\sqrt 1}=2.$$

mathlove
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  • Multiplying by the conjuagate works, no wonder. But what is wrong with doing this without conjugate multiplication, like mine? I knew the answer would be "2", but I am not asking for the solution of the problem. I am just wondering what I have done bad in the expression I wrote. Could you please tell me what is wrong with it? – user2694307 Oct 03 '14 at 15:41
  • @user2694307: I added a bit. Take a look. – mathlove Oct 03 '14 at 15:42
  • I am sorry but I couldn't understand the last line, especially the part that you turned the denominator into sqrt(1+ 2/x) + sqrt(1- 2/x). Did you divide by x^2 without multiplying? I thought it was necessary to multiply with x^2 at the same time to keep the function the same. – user2694307 Oct 03 '14 at 15:47
  • @user2694307: I added a bit again. Have a look. – mathlove Oct 03 '14 at 15:55
  • Oh, now I see it, nevermind what I asked. I must not have seen the existence of "x" in the numerator. That got me thinking what happened to the "x"s in the denominator. – user2694307 Oct 03 '14 at 15:58
  • @user2694307: Ok, glad to help you. – mathlove Oct 03 '14 at 15:59