# 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 Jun 12, 2014 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? Jun 12, 2014 at 19:32
• numerator also goes to zero. so it still need to be reduced further, as 0/0 form is indeterminate. Jun 12, 2014 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? Jun 12, 2014 at 19:36
• yes. it is possible. and then we would say that the limit is $\infty$ Jun 12, 2014 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? Jun 12, 2014 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. Jun 12, 2014 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? Jun 12, 2014 at 19:30
• @jeremyradcliff The denominator goes to $4$ in the above. Jun 12, 2014 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? Jun 12, 2014 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 $\epsilon=0$.