# Limit of $\sqrt{4x^2 + 3x} - 2x$ as $x \to \infty$

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

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.

• your denominator goes to zero – Edwin_R Jun 12 '14 at 19:28
• @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? – jeremy radcliff Jun 12 '14 at 19:32
• numerator also goes to zero. so it still need to be reduced further, as 0/0 form is indeterminate. – Edwin_R Jun 12 '14 at 19:34
• @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? – jeremy radcliff Jun 12 '14 at 19:36
• yes. it is possible. and then we would say that the limit is $\infty$ – Edwin_R Jun 12 '14 at 19:40

Hint: $\sqrt{4x^{2}+3x}-2x=\frac{3x}{\sqrt{4x^{2}+3x}+2x}=\frac{3}{\sqrt{4+\frac{3}{x}}+2}$

• 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? – jeremy radcliff Jun 12 '14 at 19:26
• 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. – user71352 Jun 12 '14 at 19:28
• 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? – jeremy radcliff Jun 12 '14 at 19:30
• @jeremyradcliff The denominator goes to $4$ in the above. – Alex G. Jun 12 '14 at 19:31
• @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? – jeremy radcliff Jun 12 '14 at 19:33

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

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

$$\sqrt{4x^2 + 3x} - 2x = \frac{\sqrt{4 + \frac{3}{x}} - 2}{1/x}$$
Hence both the numerator and denominator tend to $0$ as $x \to \infty$.
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 $\epsilon=0$.