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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.

Martin Sleziak
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jeremy radcliff
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4 Answers4

6

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

user71352
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  • 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
  • Not necessarily. $\lim_{x\to0}\frac{\sin(x)}{x}=1$ but both $\sin(x)$ and $x$ tend to $0$. Having both the numerator and denominator tend to $0$ doesn't mean the limit will not exist. – user71352 Jun 12 '14 at 19:35
  • @user71352 Thanks, is there a way to make sense of the fact that $\lim_{x\to\infty}\frac{sin(x)}{x} = 1$? Or is it just something you have to know? – jeremy radcliff Jun 12 '14 at 19:50
  • The limit should be $x$ tending to $0$ not infinity and geometrically speaking you could interpret $\sin(x)$ as an approximation to the arc of a circle when the angle, $x$, is really small so it should be that $\frac{\sin(x)}{x}\to0$ for $x\to0$. Formally you can prove the limit by noticing that $\lim_{x\to0}\frac{\sin(x)}{x}=\frac{d}{dx}(\sin(x))\vert_{x=0}$. – user71352 Jun 12 '14 at 19:55
  • @jeremyradcliff: You have asked a very genuine doubt "how do you know you need to reduce further before reaching a conclusion?". Acutally when both numerator and denominator tend to $0$ you have to understand that in this case limit could exist with any particular value or tend to $\pm\infty$ or the limit might not exist. So that there is no conclusion when you get $0/0$ and to conclude you need to do "something more". What "something more" you need to do will require some practice and experience with such problems. – Paramanand Singh Jun 14 '14 at 06:46
  • @jeremyradcliff: One can do some algebraic / trigonometric manipulation so that the earlier $0/0$ looks more like a normal calculation say $1/1$ or $3/4$. When you don't see any obvious manipulations to get rid of $0/0$ then you go for some standard limits like $\lim_{x \to 0}\dfrac{\sin x}{x} = 1$ and sometimes even this may not be easy to apply. Then you go for the high level tools like L'Hospital or Taylor's series. I have tried to explain such step by step approach to various limits in my blog http://paramanands.blogspot.com/2013/11/teach-yourself-limits-in-8-hours-part-1.html – Paramanand Singh Jun 14 '14 at 06:51
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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

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

John Alexiou
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2

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 $x \to \infty$.

Alex G.
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1

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 $\epsilon=0$.

VanillaSpinIce
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