Taking a limit and getting the wrong answer...don't know why

[SmartyPants]

TL;DR

I understand how to get to the correct answer, but the method often used is quite non-intuitive to and requires a leap of faith starting out. My method seems more intuitive (to me anyways), and I can't for the life of me understand what I'm doing wrong even though it leads to an incorrect answer.

The limit in question:

$$\lim_{x \to 0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x - \sin x} = 2$$

As stated in the TL;DR, I understand how to get to the correct answer, but I don't know that I would have figured it out without a little help/direction at the beginning because it involves adding and subtracting $tan(sin(x))$ from the numerator and having the foresight to see that this is going to lead somewhere. I started out differently based on what seems intuitive to me. Specifically, I started by multiplying both the numerator and the denominator by $\frac{1}{sin(x)tan(x)}$:

$$\lim_{x \to 0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x - \sin x}$$ $$= \lim_{x \to 0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x - \sin x}\cdot\frac{\frac{1}{sin(x)tan(x)}}{\frac{1}{sin(x)tan(x)}}$$ $$= \lim_{x \to 0} \frac{\frac{\tan(\tan x) - \sin(\sin x)}{sin(x)tan(x)}}{\frac{\tan x - \sin x}{sin(x)tan(x)}}$$ $$= \lim_{x \to 0} \frac{\frac{\tan(\tan x)}{sin(x)tan(x)}-\frac{\sin(\sin x)}{sin(x)tan(x)}}{\frac{\tan x}{sin(x)tan(x)}-\frac{\sin x}{sin(x)tan(x)}}$$ $$= \lim_{x \to 0} \frac{\frac{\tan(\tan x)}{tan(x)}\cdot\frac{1}{sin(x)}-\frac{\sin(\sin x)}{sin(x)}\cdot\frac{1}{tan(x)}}{\frac{1}{sin(x)}-\frac{1}{tan(x)}}$$ $$= \lim_{x \to 0} \frac{1\cdot\frac{1}{sin(x)}-1\cdot\frac{1}{tan(x)}}{\frac{1}{sin(x)}-\frac{1}{tan(x)}}$$ $$= \lim_{x \to 0} \frac{\frac{1}{sin(x)}-\frac{1}{tan(x)}}{\frac{1}{sin(x)}-\frac{1}{tan(x)}}$$ $$= \lim_{x \to 0} 1=1$$
Now I know the $\lim_{x \to 0} \frac{\tan(\tan x) - \sin(\sin x)}{\tan x - \sin x} = 2$. But $\frac{a}{a}$ definitely equals 1, and I don't see any algebraic mistakes or misuses of fundamental trig limits (like $\lim_{x \to 0} \frac{sin(x)}{x}=1$ and $\lim_{x \to 0} \frac{tan(x)}{x}=1$, both of which I make use of). It's gotta be something silly/stupid/trivial...can someone please spot the error?

Thanks,
Eric

 

[renormalize]

My method seems more intuitive (to me anyways), and I can't for the life of me understand what I'm doing wrong even though it leads to an incorrect answer.

Your step:$$\lim_{x\to0}\frac{\frac{\tan(\tan x)}{tan(x)}\cdot\frac{1}{sin(x)}-\frac{\sin(\sin x)}{sin(x)}\cdot\frac{1}{tan(x)}}{\frac{1}{sin(x)}-\frac{1}{tan(x)}}=\lim_{x\to0}\frac{1\cdot\frac{1}{sin(x)}-1\cdot\frac{1}{tan(x)}}{\frac{1}{sin(x)}-\frac{1}{tan(x)}}\tag{1}$$is wrong. Focus on the first term in the numerator on the left-side:$$\frac{\tan(\tan x)}{\tan x}\cdot\frac{1}{\sin x}=\frac{\tan(\tan x)}{\tan x}/\sin x\equiv f(x)/g(x)\tag{2}$$Now it's true that $\lim_{x\rightarrow0}f(x)=1$ (as you observe) but it's also true that $\lim_{x\rightarrow0}g(x)=0$, which means $\lim_{x\rightarrow0}f(x)/g(x)\neq\lim_{x\rightarrow0}1/g(x)$. The limit of a quotient is not the quotient of the limits whenever the limit of the denominator is zero:

(from https://web.ma.utexas.edu/users/m408n/AS/LM2-3-2.html)
This same observation also applies to the second term in the numerator on the left-side of eq.(1).
But what you can do instead is to apply L'Hôpital's rule to the left-side of (1) and show:$$\frac{\lim_{x\rightarrow0}\frac{d}{dx}\left(\frac{\tan(\tan x)}{tan(x)}\cdot\frac{1}{sin(x)}-\frac{\sin(\sin x)}{sin(x)}\cdot\frac{1}{tan(x)}\right)}{\lim_{x\rightarrow0}\frac{d}{dx}\left(\frac{1}{sin(x)}-\frac{1}{tan(x)}\right)}=\frac{1}{1/2}=2\tag{2}$$as expected.

 

[PeroK]

I would have been tempted to try letting $y = \sin x$. I don't know if that helps.

 

[SmartyPants]

Your step:$$\lim_{x\to0}\frac{\frac{\tan(\tan x)}{tan(x)}\cdot\frac{1}{sin(x)}-\frac{\sin(\sin x)}{sin(x)}\cdot\frac{1}{tan(x)}}{\frac{1}{sin(x)}-\frac{1}{tan(x)}}=\lim_{x\to0}\frac{1\cdot\frac{1}{sin(x)}-1\cdot\frac{1}{tan(x)}}{\frac{1}{sin(x)}-\frac{1}{tan(x)}}\tag{1}$$is wrong. Focus on the first term in the numerator on the left-side:$$\frac{\tan(\tan x)}{\tan x}\cdot\frac{1}{\sin x}=\frac{\tan(\tan x)}{\tan x}/\sin x\equiv f(x)/g(x)\tag{2}$$Now it's true that $\lim_{x\rightarrow0}f(x)=1$ (as you observe) but it's also true that $\lim_{x\rightarrow0}g(x)=0$, which means $\lim_{x\rightarrow0}f(x)/g(x)\neq\lim_{x\rightarrow0}1/g(x)$. The limit of a quotient is not the quotient of the limits whenever the limit of the denominator is zero:
View attachment 371479
(from https://web.ma.utexas.edu/users/m408n/AS/LM2-3-2.html)
This same observation also applies to the second term in the numerator on the left-side of eq.(1).
But what you can do instead is to apply L'Hôpital's rule to the left-side of (1) and show:$$\frac{\lim_{x\rightarrow0}\frac{d}{dx}\left(\frac{\tan(\tan x)}{tan(x)}\cdot\frac{1}{sin(x)}-\frac{\sin(\sin x)}{sin(x)}\cdot\frac{1}{tan(x)}\right)}{\lim_{x\rightarrow0}\frac{d}{dx}\left(\frac{1}{sin(x)}-\frac{1}{tan(x)}\right)}=\frac{1}{1/2}=2\tag{2}$$as expected.

Ah, I see. So alternatively, the limit of the product is NOT the product of the limits when either $f(x)$ or $g(x)=\frac{1}{0}=\infty$ (which is essentially the same as saying the limit of the quotient is NOT the quotient of the limits when the denominator is 0). I was trying to solve it without the help of L'Hopital's Rule, otherwise I would have applied it from the get go since we get the indeterminate form of $\frac{0}{0}$ right away...I should have mentioned that. Nevertheless, at least I understand now why my method wasn't working. Thank you.

 

[SmartyPants]

I would have been tempted to try letting $y = \sin x$. I don't know if that helps.

I did see an example of this limit being solved with the substitution you mentioned, and it does make things a bit easier to visualize.

 

[PeroK]

You can apply L'Hopital directly. Let $t(x) = \tan x$ and $s(x) = \sin x$, then:
$$t'(x) = \frac 1 {\cos^2 x}, \ t''(x) = \frac{2\sin x}{\cos^3 x}, \ t'''(x) = \frac{6\sin^2 x}{\cos^4 x} + \frac 2 {\cos^2 x}$$$$s'(x) = \cos x, \ s''(x) = -\sin x, \ s'''(x) = -\cos x$$$$t'(0) = 1, \ t''(0) = 0, \ t'''(0) = 2$$$$s'(0) = 1, \ s''(0) = 0, \ s'''(0) = -1$$So, we need to take the third derivative, in which case (with $d(x)$ as the denominator), we have:$$d'''(0) = 3$$Now for the numerator. In general, if $g(x) = f(f(x))$, we have:
$$g'''(x) = f'''(f(x))[f'(x)]^3 + 3f''(f(x))f'(x)f''(x) + f'(f(x))f'''(x)$$Applying this for $g_t(x) = \tan (\tan x)$ and $g_s(x) = \sin (\sin x)$ we get:
$$g'''_t(0) = t'''(0)t'(0) + 3t''(0)t'(0)t''(0) + t'(0)t'''(0) = 2 + 0 + 2 = 4$$$$g'''_s(0) = s'''(0)s'(0) + 3s''(0)s'(0)s''(0) + s'(0)s'''(0) = -1 + 0 - 1 = -2$$Hence, with $n(x)$ as the numerator, we have:
$$n'''(0) = 6$$And the limit is shown to be 2.

 

[PeroK]

Here's something interesting, and possibly useful:

In this case, we needed $t'''(0)$, where $t(x) = \tan x$. Note that:
$$t'(x) = \sec^2 x= 1 + \tan^2 x$$$$t''(x) = 2\tan x(1 + \tan^2x) = 2 \tan x + 2 \tan^3 x$$$$t'''(x) = (2 + 6\tan^2 x)(1 + \tan^2 x) = 2 + 8\tan^2 x + 6\tan^4 x$$And we have a simple iterative rule for generating the derivatives of $\tan x$. We can see that when $n$ is even, we have $t^{(n)}(0) = 0$. And, when $n$ is odd, $t^{(n)}(0)$ equals the leading constant term in the nth derivative.

If we wanted the fifth derivative, evaluated at 0, we can already see that this will depend only on the term in $\tan^2 x$ in the third derivative. This will generate $16\tan x$ in the 4th derivative and the constant term $16$ in the 5th derivative. So that $t^{(5)}(0) = 16$.

Note that we are effectively calculating the Taylor series coefficients for $\tan x$. And that these coefficients are related to the Bernoulli numbers:

https://en.wikipedia.org/wiki/Bernoulli_number

 

[PeroK]

Here's a solution using Taylor Series:
$$\tan x = x + \frac{x^3}{3} + \dots$$$$\sin x = x - \frac{x^3}{6} + \dots$$$$\tan x - \sin x = \frac{x^3}{2} + \dots$$$$\tan(\tan x) = (x + \frac{x^3}{3} + \dots) + \frac 1 3(x + \frac{x^3}{3} + \dots)^3 + \dots$$$$= x + \frac 2 3 x^3 + \dots$$$$\sin(\sin x) = (x - \frac{x^3}{6} + \dots) - \frac 1 6 (x - \frac{x^3}{6} + \dots)^3 + \dots$$$$= x - \frac 1 3 x^3 + \dots$$$$\tan(\tan x) - \sin(\sin x) = x^3 + \dots$$Hence:
$$\frac{\tan(\tan x) - \sin(\sin x)}{\tan x - \sin x} = \frac{x^3 + \dots}{\frac 1 2 x^3+\dots} = \frac{1 + \dots}{\frac 1 2 + \dots}$$