Removing superfluous functions from limits

Question

Suppose you have a quotient of form,

$$ \frac{ f(x) + g(x) + h(x) + \cdots}{ q(x) + r(x) + p(x)+\cdots}$$

Consider an expression of sort,

$$ \lim_{x \to a } \frac{ f(x) + g(x) + h(x) + \cdots}{ q(x) + r(x) + p(x) + \cdots}$$

Now, suppose this gives me $ \frac{0}{0}$

So, I can apply l'hopital on it maybe one or two times and I finally get some finite value of limit 'L'

that is,

$$ \lim_{x \to a } \frac{ f(x) + g(x) + h(x) + \cdots}{ q(x) + r(x) + p(x) + \cdots} =L$$

Now, suppose 'L' is invariant on removal of some of the functions (I'll illustrate with example) how do I choose functions which do not contribute to limit to simplfy calculations?

Example:

$$ \lim_{ x \to 0 } \frac{ \sin x + x^2 + x^3 + x^4 + \cdots}{x} = \lim_{ x \to 0} \frac{ \sin x }{x} $$

Now, clearly we can see that adding higher power polynomials to the limit does it not change it at all when the limit is going to 0

So, clearly limits as x goes to 0 is unaffected by polynomial terms in numerator which are have a 'degree' greater than denominator ( I don't know how to prove it but it is intuitive for me)

Now, how do I generalize this idea and be able to remove functions from limits?

Edit: I'm particularly looking for techniques for removing superflous functions from the denominator even though my illustration of what I'm saying was one in the numerator

Redit: I'm not particularly looking for a case with first derivative only, I am looking for most general thing possible

fixed it pl check the edit – tryst with freedom Jul 25 '20 at 17:02 — tryst with freedom, Jul 25 '20 at 17:02

Vincenzo Tibullo · Answer 1 · 2020-08-14T13:33:42.303

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Consider the limit $$ \lim_{x\to a}\frac{f_1(x)+\ldots+f_m(x)}{g_1(x)+\ldots+g_n(x)} $$ and suppose that $$ \lim_{x\to a}f_i(x)=\lim_{x\to a}g_j(x)=0,\qquad\forall i=1,\ldots,m,\ j=1,\ldots,n. $$ Moreover, suppose that \begin{align} &\forall i=1,\ldots,m,\ \exists h_i\in\mathbb{N}:\ \lim_{x\to a}\frac{f_i(x)}{(x-a)^{h_i}}=p_i\in\mathbb{R}-\{0\},\\ &\forall j=1,\ldots,n,\ \exists k_j\in\mathbb{N}:\ \lim_{x\to a}\frac{g_j(x)}{(x-a)^{k_j}}=q_j\in\mathbb{R}-\{0\}. \end{align} We can informally say that $f_i$ is an infinitesimal of order $h_i$ in $a$, and $g_j$ is an infinitesimal of order $k_j$ in $a$ (with respect to $x-a$).

If in the numerator there is one term, say $f_r$, such that $h_r$ < $h_i,\ \forall i=1,\ldots,m,\ i\neq r$, and the same happens in the denominator for, say, $k_s$ then we can write \begin{align} &\lim_{x\to a}\frac{f_1(x)+\ldots+f_m(x)}{g_1(x)+\ldots+g_n(x)}=\\ &\qquad=\lim_{x\to a}\frac{\dfrac{f_1(x)}{(x-a)^{h_r}}+\ldots+\dfrac{f_m(x)}{(x-a )^{h_r}}}{\dfrac{g_1(x)}{(x-a)^{k_s}}+\ldots+\dfrac{g_n(x)}{(x-a)^{k_s}}}\cdot \frac{(x-a)^{h_r}}{(x-a)^{k_s}}=(*). \end{align} We see that the in numerator of the first factor, all ratios go to $0$, except that involving $f_r$, and the same in the denominator for $g_s$, so $$ (*)=\lim_{x\to a}\frac{\dfrac{f_r(x)}{(x-a)^{h_r}}}{\dfrac{g_s(x)}{(x-a)^{k_s}}}\cdot \frac{(x-a)^{h_r}}{(x-a)^{k_s}}=\lim_{x\to a}\frac{f_r(x)}{g_s(x)}, $$ and this should answer to the question on how to eliminate unneeded terms in numerator and/or in denominator. Moreover, using previous knowledge, we can also say that $$ (*) = \frac{p_r}{q_s}\lim_{x\to a}(x-a)^{h_r-s_s}, $$ that is a limit pretty easy to calculate.

Be aware that if you have more than one term of minimal order, you cannot apply the preceeding reasoning, because the sum of terms of order $n$ can have an order higher than $n$, e.g. $$ \lim_{x\to 0}\frac{\sin(x)-x-x^2}{x^2+x^3}\neq\lim_{x\to 0}\frac{\sin(x)-x}{x^2} $$ because $\sin(x)-x$ is of order $3$, despite both $\sin(x)$ and $x$ are of order $1$, so the correct elimination is $$ \lim_{x\to 0}\frac{\sin(x)-x-x^2}{x^2+x^3}=\lim_{x\to 0}\frac{-x^2}{x^2} $$

Example

Calculate the following limit $$ \lim_{x\to0}\frac{x^2+2\sin(x^2)+3\sin^3(x)+4\tan(x)+5\log^2(x+1)}{3\cos(x)-3+\arcsin(x)+\log(1-x^2)}=(**) $$ We can observe that \begin{align} & \lim_{x\to0}\frac{x^2}{x^2}=1\neq0 \\ & \lim_{x\to0}\frac{2\sin(x^2)}{x^2}=2\lim_{t\to0}\frac{\sin(t)}{t}=2\cdot1=2\neq0 \\ & \lim_{x\to0}\frac{5\log^2(1+x)}{x^2}=5\left(\lim_{x\to0}\frac{\log(1+x)} {x}\right)^2=5\cdot1^2=5\neq0 \\ & \lim_{x\to0}\frac{3\cos(x)-3}{x^2}=-3\lim_{x\to0}\frac{1-\cos(x)}{x^2}=-3\cdot\frac{1}{2}=-\frac{3}{2}\neq0 \end{align} so all these terms are of order $2$. Next, we have \begin{align} & \lim_{x\to0}\frac{3\sin^3(x)}{x^3}=3\left(\lim_{x\to0}\frac{\sin(x)}{x}\right)^3=3\cdot1^3=3\neq0 \\ \end{align} so this term is of order $3$. Finally \begin{align} & \lim_{x\to0}\frac{4\tan(x)}{x}=4\lim_{x\to0}\frac{\tan(x)}{x}=4\cdot1=4\neq0 \\ & \lim_{x\to0}\frac{\arcsin(x)}{x}=1\neq0 \end{align} so these two terms are of order 1, the lowest in the numerator and also the lowest in the denominator, so $$ (**)=\lim_{x\to0}\frac{4\tan(x)}{\arcsin(x)} $$

edited Aug 14 '20 at 13:33

answered Jul 26 '20 at 16:03

Vincenzo Tibullo

10,955

why did you remove 0 from R? – tryst with freedom Jul 26 '20 at 16:19
and can you explain this infinitesimal order stuff? I am not very familiar with non - standard analysis – tryst with freedom Jul 26 '20 at 16:20
$0$ should be removed because otherwise we have, e.g., that $x^3$ is of order $k=0,1,2,3,$ for $x\to0$, while we only want $3$. This is not standard analysis, it is called (I translate from Italian) principle of substitution of infinitesimals, and I can only find Italian references. But try to follow the reasoning, instead of trying to enclose this in a given frame. – Vincenzo Tibullo Jul 26 '20 at 16:35
one question, what do you mean when you say order? I am trying to understand, but some terms and ideas I don't get – tryst with freedom Jul 26 '20 at 16:40
I think I would be able to understand more if you showed example of method you meant – tryst with freedom Jul 26 '20 at 16:43
what is $ h_r $ and $ k_s$ – tryst with freedom Jul 26 '20 at 16:44
The meaning of order is nothing more that what is said in the post: given the stated limit related to $f_i$, "We can informally say that $f_i$ is an infinitesimal of order $h_i$ in $a$". All try to add some examples. – Vincenzo Tibullo Jul 26 '20 at 16:46
I mean, what does it mean to have an infinitesimal of some order? – tryst with freedom Jul 26 '20 at 16:48
$f$ is an infinitesimal of order $n$ in $a$ if $$\lim_{x\to a}\frac{f(x)}{(x-a)^n}=l\neq0$$ – Vincenzo Tibullo Jul 26 '20 at 16:53
hm, I'm starting to get it I think but I think I need few example of your method – tryst with freedom Jul 26 '20 at 17:01
@DDD4C4U I added an example – Vincenzo Tibullo Jul 26 '20 at 18:00
I read this 10 times but I still can't understand the idea, can you say it in a different way? – tryst with freedom Jul 26 '20 at 18:34
why are terms of order greater than one neglected? – tryst with freedom Jul 26 '20 at 18:35
It is like $$\lim_{x\to0}\frac{x^2+x^3}{2x^2+3x^4}=\lim_{x\to0}\frac{x^2(1+x)}{x^2(2+3x^2)}=()$$ and given that $1+x\to1$ and $2+3x^2\to2$, we have $$()=\lim_{x\to0}\frac{x^2\cdot1}{x^2\cdot2}=\lim_{x\to0}\frac{x^2}{2x^2}$$ – Vincenzo Tibullo Jul 26 '20 at 18:39
I think I understand it after thinking a lot, it is like you factor out least power monomial from numerator and denominator, and then take limit – tryst with freedom Jul 26 '20 at 21:36

EuklidAlexandria · Answer 2 · 2020-07-24T16:22:12.593

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Considering your example, we have $$ \frac{\sin(x) + \sum_{j=2}^n x^j}{x} = \frac{\sin(x)}{x} + \sum_{j=1}^{n-1} x^j, $$ where the last term tends to zero for $x\to0$.

edited Jul 24 '20 at 16:22

answered Jul 24 '20 at 16:21

EuklidAlexandria

569

This was not exactly my question.. – tryst with freedom Jul 24 '20 at 16:21
@DDD4C4U If I understand you correctly, you can factor out a term in the denominator and then divide the numerator by this term as in this example. Then, apply the additivity of limits. – EuklidAlexandria Jul 24 '20 at 16:27
I didn't mean to give impression that I only wanted to reomve stuff from numerator, I want to remove from denominator as well – tryst with freedom Jul 24 '20 at 16:28
@DDD4C4U But the factored-out term (here: $x$) is removed from the denominator. Or should it be also removed from $\frac{\sin x}{x}$? – EuklidAlexandria Jul 24 '20 at 16:33
I am looking for generality , I only gave the example so that people could understand what I am saying – tryst with freedom Jul 24 '20 at 16:34
@DDD4C4U But you can generalize from this example. It is the same technique. – EuklidAlexandria Jul 24 '20 at 16:35
this additive property you've stated doesn't apply for stuff in denominators – tryst with freedom Jul 24 '20 at 16:38
@DDD4C4U Sure, but you can divide by the denominator. In this case, since the $\sin$-function is analytic, we can express it by a Taylor series and divide much further. – EuklidAlexandria Jul 24 '20 at 16:40

score 0 · Answer 3 · answered Jul 24 '20 at 16:26

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If we consider $$\frac{\sum \limits_jf_j(x)}{\sum \limits_j q_j(x)}$$ then $L=\frac{\sum \limits_jf_j'(a)}{\sum \limits_j q_j'(a)}$. So we can remove $f_k,\,q_m$ if $f_k'(a)=0$ and $q_m'(a)=0$

answered Jul 24 '20 at 16:26

Vertum

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Removing superfluous functions from limits

3 Answers3