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Could someone please show me how to derive the limit of the difference quotient of $f(x) = \frac{2}{x^2}$, as $x\rightarrow x_0$

The difference quotient is just the expression: $(f(x+h)-f(x))/h$

So, it's easy to get an expression for this, but a little tricky to eliminate the 'h' from the denominator..

Would appreciate it if someone can show me how.

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4 Answers

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The trick is simply to add the two fractions together.

$$\lim_{h \to 0} \frac{\frac{2}{(x + h)^2} - \frac{2}{x^2}}{h} = \lim_{h \to 0} \frac{\frac{2[x^2 - (x+h)^2]}{x^2(x + h)^2}}{h} = \lim_{h \to 0} \frac{\frac{2[-2hx - h^2]}{x^2(x + h)^2}}{h} = \lim_{h \to 0} \frac{2[-2x - h]}{x^2(x + h)^2} = \frac{2(-2x)}{x^4} = \frac{-4}{x^3}$$

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The answers above are good. So far, in each of them, and in your question itself, the difference quotient is taken to be $$\frac{f(x+h)-f(x)}h,\quad h\neq 0.\tag{1}$$ (The $h\neq 0$ part is important, since the expression is meaningless if $h=0$.) Now, there's no real problem with using this--and in fact, it is equivalent the alternative method I'll use here (I'll explain how after I show you my alternative approach, though it'll hopefully be clear how they're connected after some brief inspection). In this instance (and indeed, I'd say generally, when dealing with polynomials and "simple" rational functions), the method I'll demonstrate has the advantage of not having to go through the tedious process of computing binomial expansions, so long as you bear one particular trick in mind.


Your original question is about how to find the limit of the difference quotient as $x\to x_0$, and I suspect that some of your confusion stems from the fact that the $h$ term doesn't go away when you take $x\to x_0$. You could replace $x$ by $x_0$ in $(1)$ to get $$\frac{f(x_0+h)-f(x_0)}h,\quad h\neq 0,\tag{$1'$}$$ then proceed as described in the other answers to take care of the $h$ factor in the denominator. I will use instead the difference quotient $$\frac{f(x)-f(x_0)}{x-x_0},\quad x\neq x_0.\tag{2}$$ (Again, it's important that $x\neq x_0$.)

Well, the only real trick we have to remember here is that for any $a,b$, we have $a^2-b^2=(a+b)(a-b)$--the factoring formula for a difference of squares. Now, the difference quotient $(2)$ becomes $$\frac1{x-x_0}\cdot\left(\frac2{x^2}-\frac2{x_0^2}\right)=\frac1{x-x_0}\cdot\left(\frac{2x_0^2}{x_0^2x^2}-\frac{2x^2}{x_0^2x^2}\right)=\frac{2(x_0^2-x^2)}{x_0^2x^2(x-x_0)}=\frac{-2(x^2-x_0^2)}{x_0^2x^2(x-x_0)},$$ so by the difference of squares formula, $$\frac{f(x)-f(x_0)}{x-x_0}=\frac{-2(x+x_0)(x-x_0)}{x_0^2x^2(x-x_0)}=\frac{-2(x+x_0)}{x_0^2x^2}.\tag{3}$$ Now, $(3)$ holds true whenever $x_0,x,x-x_0\neq 0$--that is, whenever $x,x_0$ are distinct non-$0$ numbers. This is quite natural, since $0$ isn't in the domain of $f$--that is, $f(0)$ is not defined, so we can't let $x=0$ or $x_0=0$ in $(2)$ without it losing its meaning--and we've already restricted $x$ to values distinct from $x_0$ in the very definition of the difference quotient $(2)$. At this point, we can easily take the limit as $x\to x_0$, since we no longer have to worry about a zero denomiator. (Hover over the white space below if you'd like to check your work against mine.)

For any (fixed) non-$0$ $x_0$, we have by $(3)$ that $$\begin{equation*}\lim\limits_{x\to x_0}\cfrac{f(x)-f(x_0)}{x-x_0}=\lim\limits_{x\to x_0}\cfrac{-2(x+x_0)}{x_0^2x^2}=\cfrac{-2(x_0+x_0)}{x_0^2x_0^2}=\cfrac{-4x_0}{x_0^4}=\cfrac{-4}{x_0^3}.\end{equation*}$$

The answer achieved here is the same as by the other method. In case you've not already observed it, the trick for getting from $(1')$ to $(2)$ is to define $x:=x_0+h$, so $x\to x_0$ precisely as $h\to 0$, and $h\neq 0$ precisely when $x\neq x_0$. A similar trick will get us from $(2)$ to $(1')$. Thus, the methods are equivalent.


The Trick: Simply knowing the difference of squares formula won't be enough, usually, but the idea will be the same. First, let me introduce some abbreviating notation that you may or may not be familiar with. Given any two integers $k,n$ with $k\leq n$, and any function $g(x)$ defined on the integers from $k$ to $n$ (inclusive), we define $$\sum_{j=k}^ng(j)$$ to be the sum of $g(k), g(k+1),...,g(n)$--that is, the sum of all $g(j)$ with $j$ an integer ranging from $k$ to $n$, inclusive.

I claim that for any $a,b$ and any nonnegative integer $n$, we have $$a^{n+1}-b^{n+1}=(a-b)\sum_{j=0}^na^{n-j}b^j.\tag{#}$$ Let's look at some examples to see how that $\Sigma$ expression expands out:

$$\underline{n=0}:\quad \sum_{j=0}^0a^{0-j}b^j=1\;\;(=a^{0-0}b^0)$$

$$\underline{n=1}:\quad \sum_{j=0}^1a^{1-j}b^j=a+b\;\;(=a^{1-0}b^0+a^{1-1}b^1)$$

$$\underline{n=2}:\quad \sum_{j=0}^2a^{2-j}b^j=a^2+ab+b^2\;\;(=a^{2-0}b^0+a^{2-1}b^1+a^{2-2}b^2)$$

$$\underline{n=3}:\quad \sum_{j=0}^3a^{3-j}b^j=a^3+a^2b+ab^2+b^3\;\;(=a^{3-0}b^0+a^{3-1}b^1+a^{3-2}b^2+a^{3-3}b^3)$$

(Seeing the pattern, here? Start with $a^n$, then keep ticking the $a$'s exponent down by $1$ and the $b$'s exponent up by $1$ until you get to $b^n$.) It's easy to see that the $n=1$ case, in particular, simply gives us the difference of squares formula--which we already know works, and shows that we'll really be doing the same sort of thing even when we're dealing with integer powers greater than $2$. Now, let's see what happens in the general case when we expand the right-hand side of $(\#)$--I promise this is the only polynomial expansion we'll do: $$(a-b)\sum_{j=0}^na^{n-j}b^j=\sum_{j=0}^n(a-b)a^{n-j}b^j=\sum_{j=0}^n\left(a^{n+1-j}b^j-a^{n-j}b^{j+1}\right)$$ Now, if we rewrite the expression at the far right in the longer form, we get $$\left(a^{n+1}-a^nb\right)+\left(a^nb-a^{n-1}b^2\right)+\cdots+\left(a^2b^{n-1}-ab^n\right)+\left(ab^n-b^{n+1}\right).$$ In all (except the final) parenthesized pair, the second entry cancels with the first entry of the next pair. This gets rid of any pairs that aren't on the far left or far right, gets rid of the second entry on the far left pair, and gets rid of the first entry on the far right pair, leaving us with only $$a^{n+1}-b^{n+1},$$ as desired.


The General Case: We can easily use the trick to simplify the difference quotient of any polynomial or "simple" rational functions. By simple, I mean having form $$\frac{\alpha}{(x-\beta)^m},\quad x\neq \beta$$ for some integer $m\geq 1$ and some constants $\alpha,\beta$--in your example, we had $\alpha=2$, $\beta=0$, $m=2$. If the denominator is less nice, or if the numerator is a non-constant polynomial, we may end up having to expand some polynomial multiplication and end up doing at least as much work as if we'd just used the other method (with the $h$) in the first place.

I'll go ahead and apply it to an arbitrary simple rational function $f(x)=\frac{\alpha}{(x-\beta)^m},\: x\neq \beta$. The $m=1$ case is easy, so let's assume $m=n+1$ for some integer $n\geq 0$. Then the difference quotient is simplified using the trick, in a similar fashion to the work we did above, noting that $(x-\beta)-(x_0-\beta)=x-x_0$: $$\frac{f(x)-f(x_0)}{x-x_0}=\frac\alpha{(x-\beta)^{n+1}(x_0-\beta)^{n+1}(x-x_0)}\left((x_0-\beta)^{n+1}-(x-\beta)^{n+1}\right)=\frac{-\alpha}{(x-\beta)^{n+1}(x_0-\beta)^{n+1}}\sum_{j=0}^n(x_0-\beta)^{n-j}(x-\beta)^j,$$ and so $$\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\frac{-\alpha}{(x_0-\beta)^{2n+2}}\sum_{j=0}^n(x_0-\beta)^n.$$ (Hover over the white space below for the punchline.)

Now, we're adding up $n+1$ "$(x_0-\beta)^n$" terms in that $\Sigma$ expression (don't forget the $0$th term!), so (remembering that $m=n+1$), we get $$\begin{equation*}\lim\limits_{x\to x_0}\cfrac{f(x)-f(x_0)}{x-x_0}=\cfrac{-(n+1)\alpha (x_0-\beta)^n}{(x_0-\beta)^{2n+2}}=\cfrac{-(n+1)\alpha}{(x_0-\beta)^{n+2}}=\cfrac{-m\alpha}{(x_0-\beta)^{m+1}}.\end{equation*}$$

If you're dealing with a polynomial, instead, there are only a few differences in the approach and application. For one, you won't need to combine the expressions over a common denominator, and so that "$-$" doesn't show up. For another, it's probably easier to deal with each monomial individually, and then put it all together--that is, if (for example) $f(x)=\pi x^9-\sqrt{21}x^7+2x^2$, then we find (using the trick as above) that the respective limits (as $x\to x_0$) of difference quotients of $\pi x^9$, $-\sqrt{21}x^7$, $2x^2$ are $9\pi x_0^8$, $-7\sqrt{21}x_0^6$, $4x_0$. Then we'll just add them up to get $$\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=9\pi x_0^8-7\sqrt{21}x_0^6+4x_0.$$

We can still apply this trick when we're dealing with other rational functions--for example, partial fraction expansions put even the messiest rational function into the form of a sum of only polynomials and simple rational functions, which we'd then deal with individually and add up (in practice, partial fraction decompositions can be obnoxious to find)--but those are a bear to deal with, and I doubt you'll be asked to do difference quotient limit evaluations for those.

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+1 - absolutely fantastic top to bottom! –  Steven Stadnicki Sep 14 '12 at 23:54
    
A much-belated "thank you," Steven! I'm glad you liked it. –  Cameron Buie Sep 25 '13 at 19:16
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I tried to naively find common denominators at every step. Can you finish from here?

$$\dfrac{\frac{2}{x^2 + 2xh + h^2} - \frac{2}{x^2}}{h} = \dfrac{\frac{2x^2 - 2(x^2 + 2xh + h^2)}{x^2(x^2 + 2xh + h^2)}}{h} = \dfrac{2x^2 - 2(x^2 + 2xh + h^2)}{hx^2(x^2 + 2xh + h^2)}$$

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This answer is good, but as a side note: multiplying out the binomial is unnecessary, a little counterproductive, and a chance to make computational errors. The OP is advised to multiply things out only when it is necessary! –  rschwieb Sep 14 '12 at 16:46
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$$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{\frac{2}{(x+h)^2}-\frac{2}{x^2}}{h}=\lim_{h\to 0}\frac{-4xh-2h^2}{x^2(x+h)^2h}=$$

$$=\lim_{h\to 0}\frac{-4x-2h}{x^2(x+h)^2}=-\frac{4x}{x^4}=-\frac{4}{x^3}$$

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