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$$\lim_{x\to\infty} \sqrt{4x^2 + 3x} - 2x$$

I thought I could multiply both numerator and denominator by $\frac{1}{x}$, giving

$$\lim_{x\to\infty}\frac{\sqrt{4 + \frac{3}{x}} -2}{\frac{1}{x}}$$

then as x approaches infinity, $\frac{3}{x}$ essentially becomes zero, so we're left with 2-2 in the numerator and $\frac{1}{x}$ in the denominator, which I thought would mean that the limit is zero.

That's apparently wrong and I understand (algebraically) how to solve the problem using the conjugate, but I don't understand what's wrong about the method I tried to use.

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  • $\begingroup$ your denominator goes to zero $\endgroup$ – Edwin_R Jun 12 '14 at 19:28
  • $\begingroup$ @Edwin_R but shouldn't that indicate that the limit is undefined because of division by zero? Or is there no such thing for limits? $\endgroup$ – jeremy radcliff Jun 12 '14 at 19:32
  • $\begingroup$ numerator also goes to zero. so it still need to be reduced further, as 0/0 form is indeterminate. $\endgroup$ – Edwin_R Jun 12 '14 at 19:34
  • $\begingroup$ @Edwin_R ah ok, so a limit cannot be undefined in the same way that a normal expression like 5/0 can? What if the limit had turned out to approach 4/0, is that possible and would the limit then be undefined? $\endgroup$ – jeremy radcliff Jun 12 '14 at 19:36
  • $\begingroup$ yes. it is possible. and then we would say that the limit is $\infty$ $\endgroup$ – Edwin_R Jun 12 '14 at 19:40
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Hint: $\sqrt{4x^{2}+3x}-2x=\frac{3x}{\sqrt{4x^{2}+3x}+2x}=\frac{3}{\sqrt{4+\frac{3}{x}}+2}$

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  • $\begingroup$ I understand how to use the conjugate, I just don't understand what's wrong with the method I tried to use. I see you use multiplication by 1/x at the end, but what's wrong with using it at the beginning? $\endgroup$ – jeremy radcliff Jun 12 '14 at 19:26
  • $\begingroup$ The method you used is fine but you cannot conclude a limit from what you tried. Both the numerator and denominator will go to $0$. You will have to reduce it a bit further before a conclusion can be made. $\endgroup$ – user71352 Jun 12 '14 at 19:28
  • $\begingroup$ I see, thanks. But so, how do you know the limit isn't just undefined then since the denominator goes to zero? I mean how do you know you need to reduce more before you can reach a conclusion? $\endgroup$ – jeremy radcliff Jun 12 '14 at 19:30
  • $\begingroup$ @jeremyradcliff The denominator goes to $4$ in the above. $\endgroup$ – Alex G. Jun 12 '14 at 19:31
  • $\begingroup$ @AlexG. Sorry, I was talking about the method I used in the OP, in which both numerator and denominator tend to zero. Shouldn't that indicate an undefined limit because of the zero in the denominator? $\endgroup$ – jeremy radcliff Jun 12 '14 at 19:33
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I always change the variable with $y = \frac{1}{x}$ and take the limit to $y\rightarrow 0$

$$ {\rm limit} = \sqrt{\frac{4}{y^2} + \frac{3}{y}} - \frac{2}{y} = \left. \frac{\sqrt{3 y+4}-2}{y} \right|_{y\rightarrow 0} $$

No with LH rule

$$ {\rm limit} =\left. \frac{3}{2 \sqrt{3 y+4}} \right|_{y\rightarrow 0} = \frac{3}{4} $$

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Your method yields:

$$\sqrt{4x^2 + 3x} - 2x = \frac{\sqrt{4 + \frac{3}{x}} - 2}{1/x}$$

Hence both the numerator and denominator tend to $0$ as $x \to \infty$.

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  • 1
    $\begingroup$ For such cases, check out L'Hôpital's rule. $\endgroup$ – Garrett Jun 12 '14 at 19:30
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    $\begingroup$ @Garrett Or use the method of conjugation of the square root, as displayed below. $\endgroup$ – Alex G. Jun 12 '14 at 19:31
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Consider $x\rightarrow\infty$, then $4x^2>>3x$ and define $\epsilon=3x/4x^2<<0$ in that limit. Then $\sqrt{4x^2+3x}-2x=2x\sqrt{1+\epsilon}-2x=2x(1+\epsilon/2)+O(\epsilon^2)-2x=x\epsilon=3/4$. Where you use the expansion for $\sqrt(1+\epsilon)$ around $\epsilon=0$.

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