As a part of a limit that I was working on, I had to rewrite $$\frac{2x\tan\left(a\right)+2x}{1-x\tan\left(a\right)}$$ for $x$ near $0$. My understanding is that I can apply limit to the parts of the limit, so I did the following: $$\frac{2x\tan\left(a\right)+\lim_{x \to 0}\left(2x\right)}{1-x\tan\left(a\right)}=\frac{2x\tan\left(a\right)}{1-x\tan\left(a\right)}$$ When I plot the latter function in Desmos, its shape near $0$ is different from the shape of the original function, and consequently, my solution is incorrect. I tried to work it out more precisely and I don't understand why it doesn't work: $$ \begin{aligned} \lim_{x \to 0} \left(\frac{2x\tan\left(a\right)+2x}{1-x\tan\left(a\right)}\right) &= \lim_{x \to 0}\left(\frac{2x\tan\left(a\right)}{1-x\tan\left(a\right)}\right) + \lim_{x \to 0}\left(\frac{2x}{1-x\tan\left(a\right)}\right) \\[2pt] &= \lim_{x \to 0}\left(\frac{2x\tan\left(a\right)}{1-x\tan\left(a\right)}\right) \end{aligned} $$ However, if I apply the rule in the following way: $$ \frac{2x\tan\left(a\right)+2x}{1-\lim_{x \to 0}\left(x\tan\left(a\right)\right)} = \frac{2x\tan\left(a\right)+2x}{1}, $$ the shape of the resultant function is similar to the shape of the original function near $0$, and also my solution gives the correct result. What is the rule that allows the second step but forbids the first one?
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The limit of a sum is a sum of limits if all exist. This is a very good question! – Sean Roberson Dec 07 '24 at 01:13
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So long as $\tan(a)\not=0$ it should be obvious the limit as $x\to 0$ is $\frac01$. But the shape near $x=0$ is not as easy as that. – Henry Dec 07 '24 at 01:48
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Try factoring $x$ out of the numerator and out of the denominator. – John Wayland Bales Dec 07 '24 at 02:09
2 Answers
First of all, when approximating near $x = 0$, you do not take limit as $x \rightarrow 0$. Approximation is about neglecting the terms which don't matter, compared to the other terms in the expression. Put another way, you want to know what matters more near the point of interest. When you take limits, you are sort of looking at the end result. Limits are not about behaviour of a function in a neighbourhood, but rather about the "expected" behaviour of the function at the limit point. Your expression is: $$\frac{2x(1+\tan(a))}{1-x\tan(a)} = \frac{2x\tan(a)+2x}{1-x\tan(a)}$$
In your first attempt you deleted the second term. So, you are implying that the second term does not matter compared to the first; it is "much" smaller than the first. This is not true in general, because you do not know the value of $\tan(a)$. This will be true if $|\tan(a)| >> 1$, which means $\tan(a)+1 \approx \tan(a)$, or $x\tan(a)+x \approx x\tan(a)$. But if $|\tan(a)|$ is close to $1$, this approximation is invalid.
In your second attempt, you deleted the $-x\tan(a)$ in the denominator; here you are implying that for values of $x$ close to $0$, $|x\tan(a)| << 1$. So, the value of $x\tan(a)$ does not matter, compared to the constant $1$. This is true always as the parameter $x$ is taking the second term to $0$, but not the first.
To summarize, the approximation made in your first attempt is based on a comparison between the terms, $\tan(a)$ and $1$, while the second one on the comparison between the constant $1$ and $x\tan(a)$. You know for sure in the second one that $|x\tan(a)|$ is much smaller than $1$, for values of $x$ close to $0$.
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Thank you for your response! In my second attempt I implicitly state that x term matters in the numerator but doesn't in the denominator. From your explanation, I understand why this is a reasonable statement. Is there a rule of thumb about how to determine whether a term can be assumed to be unimportant? Let's say, is it a reasonable rule that I can't assume x to be insignificant in one term in numerator and significant in another term in numerator, but I can do that if one term is in the denominator and another is in the numerator? Thank you! – Роман Кирьянов Dec 07 '24 at 10:07
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It's not important whether the term is in the numerator or the denominator. It's important how one term affects the expression relative to the other. Eg, both x terms affect your numerator, but in the denominator, the effect of the x term will be much less compared to the constant term. – Kedar Desai Dec 07 '24 at 10:17
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Is there a rigorous rule about when I can and can't say that a function behaves like a certain function near some values? – Роман Кирьянов Dec 07 '24 at 10:25
My understanding is that I can apply limit to the parts of the limit ...
This is not always true. This only holds when both the limits exist after splitting. In particular, turning your argument into mathematic argument is same as saying
$$\lim_{x\to 0}[f(x)+g(x)]=\lim_{x\to 0}[f(x)]+\lim_{x\to 0}[g(x)]$$
If both $\displaystyle\lim_{x\to 0}[f(x)]$ and $\displaystyle\lim_{x\to 0}[g(x)]$ exists, then it's fine and your argument is correct! However, whenever they are not, then your is not correct. A common counterexample would be
$$\lim_{x\to 0}\left[\frac{-1}{x}+\frac{1}{x}\right]\ne\lim_{x\to 0}\left[\frac{-1}{x}\right]+\lim_{x\to 0}\left[\frac{1}{x}\right]$$
And in your method, I guess it should work well because both limit should go to $0$ as $x\to0$. There should not be any differences.
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