# Limits: dividing by the highest power does not work in this case

I believe to solve expressions like:

$$\lim_{x\to \infty} (2x^2 + 1)$$

we need to divide each term in the numerator and denominator by the highest power. In the previous case, the highest power is $x^2$, so we get

$$\lim_{x \to \infty} \left(2 + \frac{1}{x^2} \right) = 2+0=2$$

Now for:

$$\lim_{n \to \infty} \frac{-1}{\sqrt{n} + 2}$$

The answer given is $\frac{-1}{\infty}=0$

On the other hand, what I did was

$$\lim_{x \to \infty} \frac{-1}{1 + \frac{2}{\sqrt{n}}}=\frac{-1}{1}=-1$$

What's wrong?

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$2x^2 + 1 \neq (2+1/x^2)$, but $2x^2 + 1 = x^2 (2 + 1/x^2)$. Same thing for the second limit: you're forgetting the term you're factoring for. – Andy Mar 4 '12 at 10:54
We solve equations; we evaluate expressions. – Gerry Myerson Mar 4 '12 at 11:19
Yeah jiew be careful with the algebraic manipulation. That thinking leaded you to $-1=0$ or $1=0$ which implies every number $x=0$ . – checkmath Mar 4 '12 at 13:40
@Jiew Meng: It is useful to look before calculating. For example, it is clear that when $n$ is big, the number $\frac{-1}{\sqrt{n}+2}$ is negative but close to $0$. – André Nicolas Mar 4 '12 at 14:55

The rule you are referring to in your title "divide by the highest power" means that you divide every term in the given "rational expression" by $x^k$ where $k$ is the highest power of $x$ in the entire limit expression. This includes terms in both the numerator and the denominator. If no denominator is explicitly shown, then the denominator is 1.

For example, you could apply this rule to
$$\lim_{x\rightarrow\infty} (x^2+1)=\lim_{x\rightarrow\infty} {x^2+1\over 1}$$ to obtain $$\tag{0}\lim_{x\rightarrow\infty} {x^2+1\over1} =\lim_{x\rightarrow\infty} {1+{1\over x^2}\over{ 1\over x^2}} = \lim_{x\rightarrow\infty}x^2 (1+\textstyle{1\over x^2} ).$$ The important observation is:

Note that you are just algebraically manipulating the limit expression; but, you cannot change the limit expression, as you did when you replaced $x^2+1$ by non-equivalent expression $1+{1\over x^2}$. You can however replace $x^2+1$ by the equivalent expression (as far as taking limits is concerned) $x^2(1+{1\over x^2})$, as was done above. Dividing everything in the limit expression in the left hand side of $(0)$ by $x^2$ gives an equivalent expression because this amounts to multiplying both top and bottom by the same quantity (namely $1/x^2$).

I will also caution here that this rule is applicable in the specific scenario where

$\ \$1) you are taking a limit at infinity or minus ininifty

and

$\ \$2) the limit expression is of the form $P(x)/Q(x)$ where both $p$ and $Q$ are linear combinations of powers of $x$ and the largest power of $x$ appearing in the entire expression is nonnegative.

In this case, the rule says to: find the highest power of $x$ in the entire expression and then divide both $P(x)$ and $Q(x)$ by $x^k$. Note that this will lead to an equivalent expression since it amounts to multiplying bot top and bottom by the same quantity. Moreover, after doing this and simplifying, the limit will be easily determined.

Let's pause a second. There really is no reason to use this "divide by the highest power rule" when evaluating $$\lim_{x\rightarrow\infty} (x^2+1),$$ as this limit is clearly infinite.

However, if you had to find $$\tag{1} \lim_{x\rightarrow\infty} (x^2-x^3)$$ then things aren't so clear. Here, you could use the "divide by the highest power rule" (and note then, there is a 1 downstairs); but I think it's more clear to instead (and equivalently) factor the highest power of $x$ in the entire expression from both top and bottom (if there is one) and cancel (if you can). So for $(1)$, we would write: $$\lim_{x\rightarrow\infty} (x^2-x^3)= \lim_{x\rightarrow\infty} x^3 (\textstyle{1\over x}-1)$$ And now we can see that the limit is $-\infty$.

Or, consider the following example: $$\tag{2} \lim_{x\rightarrow\infty} {2x^3-x+2\over 3x^3-2x+1}= \lim_{x\rightarrow\infty} {x^3(2-{1\over x^2}+{2\over x^3})\over x^3( 3-{2\over x^2}+{1\over x^3})}= \lim_{x\rightarrow\infty} { 2-{1\over x^2}+{2\over x^3} \over 3-{2\over x^2}+{1\over x^3} } ={2\over3}.$$ Note that the last limit in $(2)$ could have been obtained by "dividing everything by $x^3$"; but it's of utmost importance that you divide all terms in both the numerator and denominator by $x^3$. Again, you can't change the limit expression.

For your second problem, dividing by $\sqrt n$ means you have to divide the $1$ in the numerator by $\sqrt n$ too $$\lim_{n\rightarrow\infty}{-1\over \sqrt n +2} =\lim_{n\rightarrow\infty}{-1/\sqrt n\over 1 +{2\over\sqrt n}} ={0\over 1}=0.$$ As with the first example, though, this is a bit much for this limit. The limit is clearly 0...

As mentiond by Andy, the method is better suited for something like: $$\tag{3} \lim_{n\rightarrow\infty}{-n^2\over \sqrt n +n^3}.$$ Using the "divide trick", you would divide everything by $n^3$: $$\lim_{n\rightarrow\infty}{-n^2\over \sqrt n +n^3} =\lim_{n\rightarrow\infty}{-{n^2\over n^3}\over {\sqrt n\over n^3} +1} =\lim_{n\rightarrow\infty}{-{1\over n }\over {1\over n^{5/2}} +1} ={0\over 0+1}=0.$$ Equivalently, use the "factor method" described above:

$$\lim_{n\rightarrow\infty}{-n^2\over \sqrt n +n^3} =\lim_{n\rightarrow\infty}{ n^3\cdot -{1\over n}\over n^3({1\over n^{5/2}}+1))} =\lim_{n\rightarrow\infty}{-{1\over n }\over {1\over n^{5/2}} +1} ={0\over 0+1}=0.$$

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It's usually hard for people -who are not seeing math on an every-day basis- to get that factoring is just dividing. Good job at conveying that! – Andy Mar 4 '12 at 15:54

What you are saying in your calculations is $2x^2 +1 = (2 + 1/x^2)$, which is wrong. What you want it $2x^2 +1 = x^2(2 + 1/x^2)$, thus

$$\lim_{x \to \infty} 2x^2 +1 = \lim_{x \to \infty} x^2(2+ 1/x^2) = \lim_{x\to \infty} 2x^2 = \infty$$

for your second limit you are making the same mistake: $\sqrt{n} + 2 \neq 1+ 2/\sqrt{n}$, but $\sqrt{n} + 2 = \sqrt{n}(1+2/\sqrt{n})$. Thus,

$$\lim_{x \to \infty} \frac{-1}{\sqrt{n}+2} = \lim_{x \to \infty} \frac{-1}{\sqrt{n}(1+2/\sqrt{n})} = \lim_{x \to \infty} \frac{-1}{\sqrt{n}} = 0$$

This method is usually better suited for indeterminate cases, like $\lim_{x \to \infty} \frac{2x^2 +1}{3x^2 + 4}$

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Hmm ... then what happened between line 2 of this image – Jiew Meng Mar 4 '12 at 15:08
@Jiew Meng: The numerator is $2x^2 + 4 = x^2(2+4/x^2)$ and in the denominator you factor $x^4$ inside both square roots, take it out to get $x^2$ (in both cases), then you factor again and simplify with the numerator. All this is done implicitly in your image. – Andy Mar 4 '12 at 15:17
To simplify, they divided top and bottom by $x^2$. – Mike Mar 4 '12 at 16:08