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I have a limit problem:

$\lim_{x \to \infty} \sqrt{x^2 + x} - x$

According to Wolfram|Alpha the answer is $\frac{1}{2}$

However, my calculation gives $1$. Please help me understand what I'm doing wrong. The process is:

$\lim_{x \to \infty} \sqrt{x^2 + x} - x = \lim_{x \to \infty} \sqrt{x^2(1 + \frac{1}{x})} - x = \lim_{x \to \infty} x\sqrt{1 + \frac{1}{x}} - x = \lim_{x \to \infty} x(\sqrt{1 + \frac{1}{x}} - 1)$

Let $t = \frac{1}{x}$

$\lim_{x \to 0} \frac{\sqrt{1 + t} - 1}{t} \Rightarrow (l'Hopital) \lim_{x \to 0} \frac{\frac{1}{\sqrt{1 + t}}}{1} = \lim_{x \to 0} \frac{1}{\sqrt{1+t}} = 1$

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  • $\begingroup$ You differentiated incorrectly. $\endgroup$ – copper.hat Mar 21 '14 at 14:54
  • $\begingroup$ Ah, thought it was something like that, thanks! Retaking calculus after a couple of years off, won't make the same mistake again :) $\endgroup$ – Ynhockey Mar 21 '14 at 15:21
  • $\begingroup$ math.stackexchange.com/questions/627427/… $\endgroup$ – lab bhattacharjee Mar 21 '14 at 16:22
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    $\begingroup$ @Ynhockey: Everyone maakes mistakis. $\endgroup$ – copper.hat Mar 21 '14 at 16:29
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You've lost your 2 in the derivation!

$\frac{d}{dt}\sqrt{1+t} = \frac{1}{2}\frac{1}{\sqrt{1+t}}$

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You forgot a factor $1/2$ when you took the derivative of $\sqrt{1+t}$.

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$\lim_{x \to \infty} \sqrt{x^2 + x} - x$

approach:

taking $t = \frac{1}{x}$ hence t tends to 0 So,

$\lim_{t \to 0} \sqrt{\frac{1}{t^2} + \frac{1}{t}} - \frac{1}{t}$

$= \lim_{t \to 0} \frac{\sqrt{1+t}}{t} - \frac{1}{t}$

$= \lim_{t \to 0} \frac{1}{t} (\sqrt{1+t} - 1)$

multiplying by $\frac{\sqrt{1+t} + 1}{\sqrt{1+t} + 1}$

$= \lim_{t \to 0} \frac{1}{\sqrt{1+t} + 1}$

Now apply limits

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