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

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

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

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

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.


Edit: Thanks to everyone's clear responses, I now understand:

  1. Taylor expansions of trig functions are immensely powerful in evaluating limits, in this case turning the problem into a limit of a rational function.

  2. If a factor of a limit exists and is nonzero, it can be factored out without affecting convergence of the product. Formally, suppose $\lim f(x)$ exists and $g(x)$ is a factor of $f(x)$. Then if $\lim g(x)$ exists in $\mathbb{R}\backslash\{0\}$, $\lim f(x)/g(x)$ exists also. That is to say, one can be "opportunistic" about simplifying ones limits if a factor with a real nonzero limit is discovered.

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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$$\lim_{x\to0}\frac{\sin(x)+x}x\text{ and }\lim_{x\to0}\frac{\sin(x)-x}{x^3}$$exist; they are equal to $2$ and to $-\frac16$ respectively. This explains why that decomposition is used.

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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 $x\to0$, 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 $\Bbb R\setminus\{0\}$, in case I get an indeterminate form. (That same concern motivation the previous handling of $\sin^{-2}x$.) Well, $\frac{\sin x}{x}\sim1\implies\frac{\sin x+x}{x}\sim2$, so finally the sought limit is twice $\lim_{x\to0}\frac{\sin x-x}{x^3}$.

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    $\begingroup$ Nice detailed expression of what I just gave a cursory overview of. $\endgroup$ – marty cohen Aug 21 at 20:20
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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.

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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): $ \def\lfrac#1#2{{\large\frac{#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.

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I'll try to recreate the steps.

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

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.

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

so

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

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