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Given $f$ some function such that $f=f_1+f_2+f_3+\cdots + f_n$ such that $Leading(f) = f_k$ for some $k$ which means that $f_k > f_i$ for any $i \not= k$ for all sufficiently large $x$, does $\lim \limits_{x \to \infty} f = \lim \limits_{x \to \infty} f_k$ ?!

For example : $f=2x^3-100x^2 +17x+3$ and its obvious that $Leading(f) = 2x^3$. and both approach $\infty$ as $x$ approach $\infty$.

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  • $\begingroup$ Fred's answer shows that your general 'rule' does not work, but it does work for polynomials (like your example). You have: $$\lim_{x \to \pm\infty} \left( a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x+a_0 \right) = \lim_{x \to \pm\infty} a_nx^n$$ $\endgroup$
    – StackTD
    Mar 30, 2017 at 10:55

2 Answers 2

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First of all: Your "leading" function is not well-defined. You define it as a subterm, whose limit is largest. However, this only works if $f_1,\dots,f_n$ have been previously defined and can not be rewritten to, e.g. $f_1+f_2,f_3,\dots,f_n$, etc. .
For your example this means that "Leading" might be $2x^3$, but also $1.5x^3$, etc. (basically any $cx^3$ with $c>1$).

Secondly, even if $f_1,\dots,f_n$ were fixed, this would not hold for every case. Consider the trivial example $$f=f_1+f_2=2+1$$ Then obviously,
$$\lim f_1=2>\lim f_2=1, $$ but $$2=\lim f_1\ne\lim f=3.$$

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No. Let $n=2$, $f_1(x)=\arctan x$ and $f_2=\frac{1}{2}f_1.$

Then $Leading(f)=f_1$ and $f=f_1+f_2$.

But $\lim \limits_{x \to \infty} f(x)= \frac{3 \pi}{4} \ne \frac{ \pi}{2}= \lim \limits_{x \to \infty} f_1(x)$

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