A limit I would like to find without using series and L'Hopital rule

Question

$$\lim_{x\to 0} \frac{\sin(\tan x)-\tan(\sin x)}{2 x\cos(\tan x)-2 x\cos(\sin x)+x^5}$$

I can not solve this limit without using series, and L'Hopital's Rule.

Answer 1

The beautiful thing about math is that there is no "solution with" or "solution without," only the solution, and various concordant ways of getting there. There's a technique called "eliminating cuts," mostly (this limits-without-lhopital meme notwithstanding) only useful for finding results about proofs themselves, where essentially you take a particular intermediate step, and wherever you rely on it, replace it with its proof. My first impulse was to do this with l'Hôpital's rule, first proving it with, and then using the proof to get from step to step without. I decided against this because the derivatives get very messy very quickly, and it would take seven of them to get to the point where the quotient isn't $0/0$, and the proof itself gets significantly messier if even two are needed. Taylor's theorem is essentially a consequence of l'Hôpital's rule (which, I expect, is why you don't want to use Taylor series), so an attempt to go that route without explicitly invoking Taylor series would probably get just as messy. Fortunately, however, the series of the trig functions predate Taylor's theorem by quite a long way, with Newton having found, for instance, the sine's by rearranging terms in the binomial expansion of $\sin^{-1}x = \int_0^x(1-t^2)^{-1/2}dt$. So rather than use infinite series, we'll use the same logic behind the series to come to polynomials that serve as upper and lower bounds.

So just to recap:

$$\sin^{-1}x = \int_0^x\frac{dt}{\sqrt{1-t^2}}$$ $$\tan^{-1}x = \int_0^x\frac{dt}{1+t^2}$$

These follow pretty quickly from the Cartesian equation for a semicircle and the definition of arclength as $\int_a^b\sqrt{1+(f'(x))^2}dx$ (the second isn't totally trivial, but I expect you already know these, so I won't dwell on them). You've probably noticed the absence of the arccosine. The reason for that is that both the cosines in the limit go to 1, which makes upper bounds slightly harder to establish. So let's get rid of them, rewriting the limit as

$$\lim_{x\to 0} \frac{\sin(\tan x)-\tan(\sin x)}{2x\left(\sqrt{1-\sin^2(\tan x)}-\sqrt{1-\sin^2(\sin x)}\right)+x^5}$$

Now, to break down the arctangent, we'll go right back to the very, very beginning - Elements IX.35 (the page says 36 for some reason). "If as many numbers as we please are in continued proportion, and there is subtracted from the second and last numbers equal to the first, then the excess of the second is to the first as the excess of the last is to the sum of all those before it." In modern notation (you have no idea how tempted I was to write this entire answer in the language of classical geometry), $\frac{ar-a}{a} = \frac{ar^{n+1}-a}{\sum_{k=0}^n ar^k}$, more commonly expressed

$$\sum_{k=0}^n r^k = \frac{1-r^{n+1}}{1-r}$$

Let $r = -t^2$, $n$ odd, and we've got a lower bound for $(1+t^2)^{-1}$, and as long as the value we're looking for is reasonably small (below about $17/20$ - recall Achilles and the tortoise), we can get an upper bound by doubling the last two terms. (Note that everywhere the tangent appears in the original limit it approaches zero.)

The arcsine will take a bit more finesse, but this technique can be used to put upper and lower bounds on the square of the integrand, from which a polynomial can be found whose square will be more or be less in the range in question. This time $r = +t^2$, $n$ need not be odd, and only the last term needs to be doubled provided that $t < 7/10$. Note that in both the original and the rewritten limit, the sine also always approaches zero.

So now we have these inequalities.

$$\int_0^{|x|}(1-t^2+t^4-t^6+t^8-t^{10})dt < |\tan^{-1}x| < \int_0^{|x|}(1-t^2+t^4-t^6+2t^8-2t^{10})dt$$ $$\int_0^{|x|}\left(1+\frac{t^2}{2}+\frac{3t^4}{8}+\frac{5t^6}{16}+\frac{35t^8}{128}\right)dt < \int_0^{|x|}\sqrt{1+t^2+t^4+t^6+t^8}dt < |\sin^{-1}x| < \int_0^{|x|}\sqrt{1+t^2+t^4+t^6+2t^8}dt < \int_0^{|x|}\left(1+\frac{t^2}{2}+\frac{3t^4}{8}+\frac{5t^6}{16}+\frac{99t^8}{128}\right)dt$$

The absolute value signs are needed in case $x$ is negative. You can verify for yourself that in the latter case the squares of the outermost integrands are sufficiently high and low, respectively. Evaluate the integrals; because the sine, tangent, and the outermost functions of the above inequalities are all odd functions, the absolute value signs can be eliminated by recognizing that negating x will only reverse the direction of the inequality, and since this range is symmetric, rather than explicit inequalities we'll henceforth be using an error term $\epsilon > 0$.

$$\tan^{-1}x = x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}\pm\left[\epsilon < 2\left|\frac{x^9}{9}-\frac{x^{11}}{11}\right|\right]$$ $$\sin^{-1}x = x+\frac{x^3}{6}+\frac{3x^5}{40}+\frac{5x^7}{112}\pm\left[\epsilon < \left|\frac{11x^9}{128}\right|\right]$$

So from this you can get polynomials to bound the sine and tangent themselves. Since both the sine and tangent are increasing at zero, the lower bounds of the functions can be found from the lower bounds of the inverse and vice versa.

For sufficiently small $x$ and $y$ (defined here as small enough that the highest-degree terms can be consolidated - remember, these are finite, so there will always be such a range),

$$y = x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}\pm\left[\epsilon < 2\left|\frac{x^9}{9}-\frac{x^{11}}{11}\right|\right] \implies x=y+\frac{y^3}{3}+\frac{2y^5}{15}+\frac{17y^7}{315}\pm[\epsilon<|cy^9|]$$ $$y = x+\frac{x^3}{6}+\frac{3x^5}{40}+\frac{5x^7}{112}\pm\left[\epsilon < \left|\frac{11x^9}{128}\right|\right] \implies x = y-\frac{y^3}{6}+\frac{y^5}{120}-\frac{y^7}{5040}\pm[\epsilon<|cy^9|]$$

For some $c$ independent of $x$. You can convince yourself of these with high school algebra if you really must, but I warn you it'll be tedious. So first, the numerator. This'll look familiar - it's exactly as if we'd used Taylor's theorem.

$$\sin(\tan x) = x + \left(\frac{1}{3}-\frac{1}{6}\right)x^3 + \left(\frac{2}{15}-\frac{1}{6}+\frac{1}{120}\right)x^5 + \left(\frac{17}{315}-\frac{11}{90}+\frac{1}{72}-\frac{1}{5040}\right)x^7 \pm [\epsilon<|cx^9|] = x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{55x^7}{1008}\pm [\epsilon<|cx^9|]$$ $$\tan(\sin x) = x + \left(-\frac{1}{6}+\frac{1}{3}\right)x^3 + \left(\frac{1}{120}-\frac{1}{6}+\frac{2}{15}\right)x^5 + \left(-\frac{1}{5040}+\frac{13}{360}-\frac{1}{9}+\frac{17}{315}\right)x^7 \pm [\epsilon<|cx^9|] = x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{107x^7}{5040} \pm [\epsilon<|cx^9|]$$ $$\sin(\tan x) - \tan(\sin x) = -\frac{x^7}{30} \pm [\epsilon<|cx^9|]$$

The denominator will be trickier. First, let's work out

$$\sin(\sin x) = x + \left(-\frac{1}{6}-\frac{1}{6}\right)x^3 + \left(\frac{1}{120}+\frac{1}{12}+\frac{1}{120}\right)x^5 + \left(-\frac{1}{5040}-\frac{13}{720}-\frac{1}{144}-\frac{1}{5040}\right)x^7 \pm [\epsilon<|cx^9|] = x - \frac{x^3}{3} + \frac{x^5}{10} - \frac{8x^7}{315} \pm [\epsilon<|cx^9|]$$

So

$$\sin^2(\sin x) = x^2 + \left(2\cdot-\frac{1}{3}\right)x^4 + \left(\frac{1}{9}+2\cdot\frac{1}{10}\right)x^6 \pm [\epsilon<|cx^8|] = x^2 - \frac{2}{3}x^4 + \frac{14}{45}x^6 \pm [\epsilon<|cx^8|]$$ $$\sin^2(\tan x) = x^2 + \left(2\cdot\frac{1}{6}\right)x^4 + \left(\frac{1}{36}+2\cdot-\frac{1}{40}\right)x^6 \pm [\epsilon<|cx^8|] = x^2 + \frac{1}{3}x^4 - \frac{1}{45}x^6 \pm [\epsilon<|cx^8|]$$

And from there

$$1 - \sin^2(\sin x) = 1 - x^2 + \frac{2}{3}x^4 - \frac{14}{45}x^6 \pm [\epsilon<|cx^8|] = \left(1 - \frac{1}{2}x^2 + \frac{5}{24}x^4 - \frac{37}{720}x^6 \pm [\epsilon<|cx^8|]\right)^2$$ $$1 - \sin^2(\tan x) = 1 - x^2 - \frac{1}{3}x^4 + \frac{1}{45}x^6 \pm [\epsilon<|cx^8|] = \left(1 - \frac{1}{2}x^2 - \frac{7}{24}x^4 - \frac{97}{720}x^6 \pm [\epsilon<|cx^8|]\right)^2$$ $$2x\left(\sqrt{1-\sin^2(\tan x)}-\sqrt{1-\sin^2(\sin x)}\right)+x^5 = x^5 + \left(2x - x^3 - \frac{7}{12}x^5 - \frac{97}{360}x^7\right) - \left(2x - x^3 + \frac{5}{12}x^5 - \frac{37}{360}x^7\right) \pm [\epsilon<|cx^9|] = x^5 - x^5 - \frac{1}{6}x^7 \pm [\epsilon<|cx^9|]$$

So you end up with

$$\lim_{x\to0}\frac{-\frac{1}{30}x^7 \pm [\epsilon<|cx^9|]}{-\frac{1}{6}x^7 \pm [\epsilon<|cx^9|]} = \frac{1}{5}$$