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I am going over my test and I am not sure what I was supposed to do for this question.

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

I attempted to do the algebra but my algebra skills are too weak to do it so I resorted to logic. I stated that the top will reduce to a positive x and the bottom will reduce to a positive $2x$ giving $ \frac {x}{2x}$ which will then give me $1/2$ but this answer is wrong. Apparently the answer is correct but my reasoning is wrong.

I don't know why it is wrong but it is incredibly frustrating to me that I can get the correct answer with logical conclusions but still get the question on the test wrong. This is my second time taking calculus and at this rate I am going to fail again, no matter how hard I try I just can't get it right for some reason.

I guess this could be a broader question, but how do you take a math test? It is a mystery to me and I have absolutely no idea what a teacher expects on tests. They showed us this method in class, but on a test it is incorrect. To further complicate things I got the epsilon delta problem wrong because I didn't show the absolute value and all that stuff at the start even though I was able to show the correct epsilon.

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Don't give up! Also, the "this could be a broader question..." IS a broader question - too much so for this site, in fact. If you ever want to discuss such things, maybe we could do it in chat. Ping me. –  The Chaz 2.0 Feb 14 '12 at 3:39
    
You do have to be careful with your ad-hoc reasoning. For instance, $\lim_{x\rightarrow\infty} [\sqrt{x^2+x}-x]=\frac12$. You might be inclined to believe this limit "reduces" to $x-x=0$. –  dls Feb 14 '12 at 4:10

3 Answers 3

up vote 6 down vote accepted

Your intuition is actually correct. Most likely, your teacher wanted the 'proper' steps and justification for that intuition. To be fair to your teacher, intuition might not always be right, and without proper mathematical justification, what you write is just English :-)

We can rewrite as

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

as $\frac{3}{y} \to 0$ as $y \to \infty$.

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+1 Beat me to it! :) –  user21436 Feb 14 '12 at 2:09

A slightly different way to think about this problem:

If you have an expression of the form where you have a polynomial raised to a fractional power divided by another polynomial raised to a fractional power, such as $$ \sqrt{x^2-3x}\over 3-2x $$ and you're taking a limit at infinity or negative infinity: $$\tag{1} \lim_{x\rightarrow-\infty} {\sqrt{x^2-3x}\over 3-2x}. $$ Then identify the "highest power of $x$" that you see taking into account any outer fractional powers. I'll call this the dominant $x$ term of the expression. This will be the larger of the dominant terms of the numerator and denominator. If the dominant terms are the same in the numerator and denominator, they are the dominant term for the entire expression.

Back to our limit $(1)$:

In the denominator, the dominant $x$ term is $x$; in the numerator the dominant $x$ term is $\sqrt {x^2}=x$.

So, the dominant $x$ term of the entire expression is just $x$.

Now divide both top and bottom of the limit expression by $x$. Note that $x$ is negative since we are taking the limit at $-\infty$; so, we have to be careful when bringing $x$ into the radical:

$$ {\sqrt{x^2-3x}\over 3-2x} ={{\sqrt{x^2-3x}\over x}\over {3-2x\over x}} ={-{\sqrt {x^2-3x \over x^2 }}\over {{3\over x}-2}} ={-{\sqrt {1-{3\over x} }}\over {{3\over x}-2}}. $$ Then take the limit: $$ \lim_{x\rightarrow-\infty} {\sqrt{x^2-3x}\over 3-2x} =\lim_{x\rightarrow-\infty}{-{\sqrt {1-{3\over x} }}\over {{3\over x}-2}}= {1\over 2}. $$


This is entirely rigorous and always works.


The process is particularly simple when taking a limit at infinity of a rational expression:

For example:

$$ \eqalign{ \lim_{x\rightarrow\infty} {x^2-2x^7-1\over 3x^2+2x+3x^7} &=\lim_{x\rightarrow\infty} { {x^2\over x^7}-{2x^7\over x^7}-{1\over x^7} \over {3x^2\over x^7}+{2x\over x^7}+{3x^7\over x^7}}\cr &=\lim_{x\rightarrow\infty} { {1\over x^5}-{2 }-{1\over x^7} \over {3 \over x^5}+{2 \over x^6}+{3}}\cr &=-{2\over3}. } $$

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It's instructive to explicitly point out that the method employed by Aryabhata (and elaborated in David's answer) is just a special case of computing limits by employing power series expansion, using Taylor/Laurent/Puiseux series. E.g the transformation $\sqrt{x^2-3x} = x \sqrt{1-3/x}$ to obtain the "dominant term" $x$ amounts to computing the leading term in the Laurent series at $x = +\infty$

$$ (x^2 - 3x)^{1/2}\: =\ x (1-3/x)^{1/2}\: =\ x -\frac{3}2 - \frac{9}8 x^{-1} - \frac{27}{16} x^{-2} + \:\cdots$$

or, perhaps more familiar, in terms of $\:z = 1/x\:$ at $\:z = 0^{+}$

$$ (z^{-2}-3z^{-1})^{1/2}\: =\ \left(\frac{1-3z}{z^2}\right)^{1/2} =\ \frac{1}z - \frac{3}2 -\frac{9}8 z - \frac{27}{16} z^2 +\:\cdots$$

Note that the limit of a Laurent series $f$ is simply the limit of its dominant leading term ${\rm L}(f)$

$$ \lim_{z\:\to\: 0^{+}}\: (a z^n + b z^{n+1} + c z^{n+2}+\:\cdots\:)\ = \lim_{z\:\to\: 0^{+}} a z^n\: (1 + \frac{b}a z + \frac{c}a z^2 + \:\cdots)\ = \lim_{z\:\to\: 0^{+}} az^n$$

Thus for purposes of computing limits we need only compute the leading term of the series.

Further, since the leading term map is multiplicative ${\rm L}(f/g) = {\rm L}(f)/{\rm L}(g)$, to compute the limit of a quotient it suffices to compute the limit of the quotient of their leading terms. This explains the algebra behind the heuristics in the other answers. The conceptual background will become clearer when one studies (higher-rank) valuation theory. In fact, employing such ideas yields effective algorithms for computing limits of a large class of elementary functions. To learn more google "transseries" and see this answer.

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