# Removing superfluous functions from limits

Mathematics Asked on January 1, 2022

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

Consider the limit $$lim_{xto a}frac{f_1(x)+ldots+f_m(x)}{g_1(x)+ldots+g_n(x)}$$ and suppose that $$lim_{xto a}f_i(x)=lim_{xto a}g_j(x)=0,qquadforall i=1,ldots,m, j=1,ldots,n.$$ Moreover, suppose that begin{align} &forall i=1,ldots,m, exists h_iinmathbb{N}: lim_{xto a}frac{f_i(x)}{(x-a)^{h_i}}=p_iinmathbb{R}-{0},\ &forall j=1,ldots,n, exists k_jinmathbb{N}: lim_{xto a}frac{g_j(x)}{(x-a)^{k_j}}=q_jinmathbb{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 $$h_r$$ < $$h_i, forall i=1,ldots,m, ineq r$$, and the same happens in the denominator for, say, $$k_s$$ then we can write begin{align} &lim_{xto a}frac{f_1(x)+ldots+f_m(x)}{g_1(x)+ldots+g_n(x)}=\ &qquad=lim_{xto 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 $$(*)=lim_{xto a}frac{dfrac{f_r(x)}{(x-a)^{h_r}}}{dfrac{g_s(x)}{(x-a)^{k_s}}}cdot frac{(x-a)^{h_r}}{(x-a)^{k_s}}=lim_{xto a}frac{f_r(x)}{g_s(x)},$$ 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 $$(*) = frac{p_r}{q_s}lim_{xto a}(x-a)^{h_r-s_s},$$ 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. $$lim_{xto 0}frac{sin(x)-x-x^2}{x^2+x^3}neqlim_{xto 0}frac{sin(x)-x}{x^2}$$ because $$sin(x)-x$$ is of order $$3$$, despite both $$sin(x)$$ and $$x$$ are of order $$1$$, so the correct elimination is $$lim_{xto 0}frac{sin(x)-x-x^2}{x^2+x^3}=lim_{xto 0}frac{-x^2}{x^2}$$

Example

Calculate the following limit $$lim_{xto0}frac{x^2+2sin(x^2)+3sin^3(x)+4tan(x)+5log^2(x+1)}{3cos(x)-3+arcsin(x)+log(1-x^2)}=(**)$$ We can observe that begin{align} & lim_{xto0}frac{x^2}{x^2}=1neq0 \ & lim_{xto0}frac{2sin(x^2)}{x^2}=2lim_{tto0}frac{sin(t)}{t}=2cdot1=2neq0 \ & lim_{xto0}frac{5log^2(1+x)}{x^2}=5left(lim_{xto0}frac{log(1+x)} {x}right)^2=5cdot1^2=5neq0 \ & lim_{xto0}frac{3cos(x)-3}{x^2}=-3lim_{xto0}frac{1-cos(x)}{x^2}=-3cdotfrac{1}{2}=-frac{3}{2}neq0 end{align} so all these terms are of order $$2$$. Next, we have begin{align} & lim_{xto0}frac{3sin^3(x)}{x^3}=3left(lim_{xto0}frac{sin(x)}{x}right)^3=3cdot1^3=3neq0 \ end{align} so this term is of order $$3$$. Finally begin{align} & lim_{xto0}frac{4tan(x)}{x}=4lim_{xto0}frac{tan(x)}{x}=4cdot1=4neq0 \ & lim_{xto0}frac{arcsin(x)}{x}=1neq0 end{align} so these two terms are of order 1, the lowest in the numerator and also the lowest in the denominator, so $$(**)=lim_{xto0}frac{4tan(x)}{arcsin(x)}$$

Answered by enzotib on January 1, 2022

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 by Vertum on January 1, 2022

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 $$xto0$$.

Answered by EuklidAlexandria on January 1, 2022

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