Mdthbs

I'm reviewing for the Math GRE Subject test and came across this question in the excellent UCLA notes.

$$lim_{xto 0}left(frac{1}{x^2}-frac{1}{sin^2 x}right).$$

If one attacks this with naive applications of L'Hospital after combining fractions, it quickly gets out of hand.
The suggested solution involves factoring it as

$$frac{1}{x^2}-frac{1}{sin^2 x} = frac{sin^2 x - x^2}{x^2sin^2 x} = left(frac{x^2}{sin^2 x}right)left(frac{sin x + x}{x}right)left(frac{sin x - x}{x^{3}}right),$$

where indeed each factor has a real positive limit.
I am wondering: What intuition or thought process might lead me to this particular factorization? Such a factorization seems non-obvious to me as it requires the introduction of another factor of $x^2$ into the numerator.

Since$$sin(x)=x-frac{x^3}{3!}+frac{x^5}{5!}-cdots,$$you know that$$sin(x)+x=2x-frac{x^3}{3!}+frac{x^5}{5!}-cdots$$and that$$sin(x)-x=-frac{x^3}{3!}+frac{x^5}{5!}-cdots.$$Therefore both limits$$lim_{xto0}frac{sin(x)+x}xtext{ and }lim_{xto0}frac{sin(x)-x}{x^3}$$exist; they are equal to $2$ and to $-frac16$ respectively. This explains why that decomposition is used.

OK, let's talk through the thought process.

• When I see a difference of two fractions, I give them common
  denominators first, as per your second $=$.




• I can't help but factorise the new numerator's difference of two
  squares after that.




• Since there's a division by $sin^2 x$, which $to0$ as $xto0$, I
  need to take out a $left(frac{sin x}{x}right)^{-2}$ factor before
  I can go any further.




• That leaves me with $frac{(sin x+x)(sin x-x)}{x^4}$. There are any
  number of ways I could write that as a product of two other factors,
  but I don't want to factor out a limit outside $Bbb
  Rsetminus{0}$, in case I get an indeterminate form. (That same
  concern motivation the previous handling of $sin^{-2}x$.) Well,
  $frac{sin x}{x}sim1impliesfrac{sin x+x}{x}sim2$, so finally
  the sought limit is twice $lim_{xto0}frac{sin x-x}{x^3}$.

Once you know that
$dfrac{sin x}{x}
to 1$,
and similar,
fairly simple limits,
it becomes natural
to try to make them appear
by extracting them
from more complicated expressions.

I'll try to recreate the steps.

Step 1): We are subtracting two fractions so let's have a common denominator, leading to
$$lim_{x rightarrow 0}{sin^2x - x^2 over x^2sin^2 x}$$

Step 2): I see a difference of two squares so let's factor:
$$lim_{x rightarrow 0}{(sin x - x)(sin x + x) over x^2 sin^2 x}$$
Step 3): I know the limit of ${sin x over x}$ is $1$, so I'll stick an $x^2$ in the numerator to place over the $sin^2 x$. The result is
$$lim_{x rightarrow 0}{x^2 over sin^2 x}{(sin x - x)(sin x + x) over x^4}$$
Step 4): I know that $sin x - x$ is $O(x^3)$ as $x rightarrow 0$, and $sin x + x$ is $O(x)$ as $x rightarrow 0$. This suggests writing the limit as
$$lim_{x rightarrow 0}{x^2 over sin^2 x}{(sin x - x) over x^3}{(sin x + x) over x}$$
And there we have it.

$$
frac {1}{x^2}-frac{1}{sin^2x} = left(frac {1}{x}+frac{1}{sin x}right)left(frac {1}{x}-frac{1}{sin x}right) = frac{1}{x^2}left(frac {x}{x}+frac{x}{sin x}right)left(frac {x}{x}-frac{x}{sin x}right)
$$

so

$$
lim_{xto 0}left(frac {1}{x^2}-frac{1}{sin^2x}right) =2lim_{xto 0}frac{1}{x^2}left(1-frac{x}{sin x}right) = 2lim_{xto 0}frac{x}{sin x}frac{1}{x^2}left(frac{sin x}{x}-1right) = 2lim_{xto 0}frac{1}{x^2}left(frac{sin x}{x}-1right) = 2lim_{xto 0}frac{1}{x^2}left(1-frac{x^2}{3!}+o(x^4)-1right) = -frac 13
$$

Since nobody has explicitly mentioned it, I am going to state for the record that the best way to evaluate limits in general is not actually via purely algebraic manipulation of the kind you are looking for, but rather via asymptotic expansions (which is how all computer algebra systems do it):
$
deflfrac#1#2{{largefrac{#1}{#2}}}
$

As $x → 0$:

  $sin(x) ∈ x - lfrac{x^3}{6} + O(x^5)$.

  Thus $lfrac1{x^2} - lfrac1{sin(x)^2} ∈ lfrac1{x^2} - lfrac1{(x - lfrac{x^3}{6} + O(x^5))^2} = lfrac1{x^2} - lfrac1{x^2·(1 - lfrac{x^2}{3} + O(x^4))}$

    $ ⊆ lfrac1{x^2} - lfrac{1+lfrac{x^2}{3}+O(x^4)}{x^2} ⊆ -lfrac13 + O(x^2)$

Now you may ask, how did I know how much of the expansion of $sin$ I would need? Well, I did not have to know. If you expand too little, you will obtain too loose asymptotic bounds at the end, and you can trace to see which expansions you need a tighter bound on. See here for more detailed explanation and examples.

Furthermore, we can sometimes utilize asymptotic expansion to find a purely algebraic proof, such as for this limit. Specifically, if $f(x) gg g(x)$, then $lfrac1{f(x)} - lfrac1{f(x)+g(x)}$ $= lfrac{g(x)}{f(x)·(f(x)+g(x))}$ $sim lfrac{g(x)}{f(x)^2}$, where $A sim B$ denotes that $A-B ll A,B$ (i.e. the discrepancy is negligible compared to the terms themselves). As a difference, the most significant terms may cancel, but once combined we will no longer have such cancellation. That is why we have to combine in this case:

$lfrac1{x^2} - lfrac1{sin(x)^2} = lfrac{sin(x)^2-x^2}{x^2·sin(x)^2}$.

And by the same asymptotic reasoning the next step is also well-motivated by the fact that as $x → 0$, we have $sin(x)^2 in x^2·(1+o(1))$, equivalently $sin(x)^2/x^2 ≈ 1$ and often denoted as $sin(x)^2 sim x^2$. So we can without thinking do:

$lfrac{sin(x)^2-x^2}{x^2·sin(x)^2} = lfrac{x^2}{sin(x)^2} · lfrac{sin(x)^2-x^2}{x^4}$

Even if we do not see the factorization, we can proceed systematically in the same fashion of factoring out significant terms:

$lfrac{sin(x)^2-x^2}{x^4} = lfrac{x^2·(1+(sin(x)-x)/x)^2-x^2}{x^4}$

And from there everything is straightforward.