Getting incorrect result on limit of square root

Question

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

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

Ah, thought it was something like that, thanks! Retaking calculus after a couple of years off, won't make the same mistake again :) — Ynhockey, Mar 21 '14 at 15:21
http://math.stackexchange.com/questions/627427/puzzled-by-displaystyle-lim-x-to-infty-sqrtx2x-x — lab bhattacharjee, Mar 21 '14 at 16:22

score 5 · Answer 1 · answered Mar 21 '14 at 14:53

5

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

answered Mar 21 '14 at 14:53

Robert Israel

448,999

score 2 · Accepted Answer · answered Mar 21 '14 at 14:54

2

You've lost your 2 in the derivation!

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

answered Mar 21 '14 at 14:54

nelv

1,310
8
20

Ankit Haldar · Answer 3 · 2014-03-21T19:12:23.093

0

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

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

edited Mar 21 '14 at 19:12

answered Mar 21 '14 at 19:03

Ankit Haldar

1

Getting incorrect result on limit of square root

3 Answers3