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.