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Question: calculate:

$$\lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + 1}$$

Attempt at a solution:

This can be written as:

$$\lim_{x \to -\infty} \frac{3 + \frac{1}{x}}{\sqrt{1 + \frac{3}{x}} + \sqrt{1 + \frac{1}{x^2}}}$$

Here we can clearly see that if x would go to $+\infty$ the limit would converge towards $\frac{3}{2}$. But what happens when x goes to $-\infty$.

From the expression above it would seem that the answer would still be $\frac{3}{2}$. My textbook says it would be $- \frac{3}{2}$ and I can't understand why.

I am not supposed to use l'Hospital's rule for this exercise.

Community
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Lukas Arvidsson
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  • You should put it in the form where you can apply L'Hospital and see if that's what it's really equal to, just the same way they determined the other limit is $+3/2$. – Gregory Grant Mar 08 '16 at 14:13
  • Hi Gregory, I am not supposed to use the L'Hospital rule for this exercise. But thank you for your comment. – Lukas Arvidsson Mar 08 '16 at 14:14
  • See http://math.stackexchange.com/questions/1607773/trick-to-solve-this-limit-lim-x-to-infty-sqrtx23x-x/1607815#1607815 – lab bhattacharjee Mar 08 '16 at 14:20

4 Answers4

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In fact:

$\lim_{x \to -\infty} \sqrt{x^2 + 3x} - \sqrt{x^2 + 1}$

$ = \lim_{x \to -\infty} \frac{x*(3 + \frac{1}{x})}{(-x)(\sqrt{1 + \frac{3}{x}} + \sqrt{1 + \frac{1}{x^2}})}$

$ = \lim_{x \to -\infty} \frac{-3 - \frac{1}{x}}{\sqrt{1 + \frac{3}{x}} + \sqrt{1 + \frac{1}{x^2}}} = -\frac{3}{2}$

crbah
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The reason your sign has changed from what it should be, is you illegally pulled something out of the square roots on the denominator.

$$\sqrt{a^2b}=|a|\sqrt{b}$$

the absolute value sign being essential.

goonfiend
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$$\sqrt{x^2+3x}-\sqrt{x^2+1}=\frac{3x-1}{\sqrt{x^2+3x}+\sqrt{x^2+1}}\cdot\frac{\frac1{-x}}{\frac1{-x}}=$$

$$=\frac{-3+\frac1x}{\sqrt{1+\frac3x}+\sqrt{1+\frac1{x^2}}}\xrightarrow[x\to-\infty]{}-\frac32$$

You need the $\;-x\;$ in order to make it positive and then you can take it into the square root.

DonAntonio
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  • The crucial point is that $\frac{1}{x}$ goes to 0 whether x goes to positive infinity or negative infinity. – user247327 Mar 08 '16 at 14:33
  • @user247327 Thank you. Yes, of course. At this point of basic calculus I think that must already be trivial. – DonAntonio Mar 08 '16 at 14:34
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$$\lim_{x\to-\infty}\sqrt{x^2+3x}-\sqrt{x^2+1}=\lim_{x\to-\infty}\left(\sqrt{x^2+3x}-\sqrt{x^2+1}\right)\cdot 1$$ $$=\lim_{x\to-\infty}\sqrt{x^2+3x}-\sqrt{x^2+1}\cdot\frac{\sqrt{x^2+3x}+\sqrt{x^2+1}}{\sqrt{x^2+3x}+\sqrt{x^2+1}}$$ $$=\lim_{x\to-\infty}\frac{\sqrt{x^2+3x}^2-\sqrt{x^2+1}^2}{\sqrt{x^2+3x}+\sqrt{x^2+1}}=\lim_{x\to-\infty}\frac{x^2+3x-(x^2+1)}{\sqrt{x^2\left(1+\frac{3}{x}\right)}+\sqrt{x^2\left(1+\frac{1}{x}\right)}}$$ $$=\lim_{x\to-\infty}\frac{3x-1}{2x}=-\frac{3}{2}$$

Madrit Zhaku
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