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Suppose $\lim\limits_{x\to\infty} f(x) = L$ and $\lim\limits_{x\to\infty}( g(x)-f(x))=0$. How to show that $\lim\limits_{x\to\infty} g(x)=L$?

My attempt:

using the Cauchy's definition of limit, we can tell that for any $\varepsilon_f > 0$ there exists $x_f$ such that for all $x>x_f$ we have $|f(x)-L| <\varepsilon_f$.

For the second limit, there's $x_g$ such that for all $x>x_g$ we have $|g(x)-f(x)| <\varepsilon_g$.

How can we combine both to conclude that $\lim\limits_{x\to\infty} g(x)=L$? Adding both inequalities:

$|f(x)-L|+|g(x) - f(x)|<\varepsilon_f+\varepsilon_g$

Since there's an absolute value on the left side, the $f(x)$ won't cancel out, so I'm not sure what to do next.

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    $\begingroup$ What do you mean by $\lim_{l \to \infty} g(x) = f(x)$? $\endgroup$
    – user296602
    Dec 29 '15 at 19:53
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    $\begingroup$ Ok, second question: What do you mean by $\lim_{x \to \infty} g(x) = f(x)$? $\endgroup$
    – user296602
    Dec 29 '15 at 19:55
  • $\begingroup$ Probably that the functions become indistinguishable in the limit. What's the correct notation? $\endgroup$ Dec 29 '15 at 19:57
  • $\begingroup$ @user5539357 Do you mean that $\lim (f(x)-g(x)) = 0$? $\endgroup$
    – Scounged
    Dec 29 '15 at 19:57
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    $\begingroup$ The best thing you will learn here is not the conclusion, but that the statement $\lim g(x) = f(x)$ makes no sense. $\endgroup$
    – zhw.
    Dec 29 '15 at 19:58
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You are going on the right way. Here it is a solution:

Let $\epsilon > 0.$ Consider $\varepsilon_f = \varepsilon_g = \varepsilon /2.$

Let $A= \max\{x_f,x_g \}$. For $x > A \geq x_f,x_g$ we have

$$ |g(x) - L| \leq |g(x) - f(x)| + |f(x) - L| < \varepsilon /2 + \varepsilon /2 = \varepsilon .$$

Then the limit of $g$ as $x\rightarrow \infty$ is $L.$

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This answer may not be very instructive, but it is another case of how to prove something in analysis with a simple trick:

Assuming linearity of limits (a result typically shown the first time talking about limits), $$\lim_{x\rightarrow\infty}g(x)=\lim_{x\rightarrow\infty}(g(x)-f(x)+f(x))=\lim_{x\rightarrow\infty}(g(x)-f(x))+\lim_{x\rightarrow\infty}f(x)=0+L.$$

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  • $\begingroup$ Adding 0 is a common trick in analysis. $\endgroup$ Dec 29 '15 at 20:08
  • $\begingroup$ There is a little problem in your solution: it is right if you suppose the existence of the limit of g. But considering the existence it is perfect $\endgroup$ Dec 29 '15 at 20:11
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Since $\lim_{x\to \infty}f(x)=L$, and $\lim_{x\to\infty}(g(x)-f(x))=0$, given $\varepsilon>0$, there exists some $x_{\varepsilon}>0$ such that $$ \max\{|L-f(x)|,|g(x)-f(x)|\}\le \frac12\varepsilon \quad (\forall x>x_\varepsilon). $$ Hence, for $x>x_\varepsilon$ we have: $$ |L-g(x)|=|L-f(x)+f(x)-g(x)|\le |L-f(x)|+|f(x)-g(x)|\le \frac12\varepsilon+\frac12\varepsilon=\varepsilon, $$ i.e. $\lim_{x\to \infty}g(x)=L$.

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