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If for a function a finite or infinite double limit $$ \lim\limits_{(x,y)\rightarrow(a,b)} f(x,y)$$ exists, and if for any $y \in Y$ there is a finite limit $$ \varphi(y) = \lim\limits_{x \rightarrow a}f(x,y) $$ then the repeated limit $$ \lim\limits_{y \rightarrow b}\varphi(y) = \lim\limits_{y \rightarrow b}\lim\limits_{x \rightarrow a}f(x,y) $$ exists and is equal to the double limit of the function.

According the definition of double limit, if given a $\varepsilon > 0$, then we can find a $\delta >0$, let $|x - a| < \delta$ and $|y -b| < \delta$, then $|f(x,y) - A| < \varepsilon$. Now fix $y$ and let it satisfies $|y - b| < \varepsilon$, then how can i replace $f(x,y)$ with $\varphi(y)$ for the inequality $|f(x,y) - A| < \varepsilon$ ? Can i reduce the domain of $x$ to $|x - a| < \varepsilon$, then calculate the limit of $f(x,y)$ from $x \rightarrow a$?

maplgebra
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1 Answers1

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By the reverse triangle inequality, we have

$$|\, |f(x,y)- A| - |\varphi(y) - A| \, | \leqslant |f(x,y) - \varphi(y)|,$$

and since $\lim_{x \to a}f(x,y) \to \varphi(y)$ it follows that $\lim_{x \to a}|f(x,y)- A| = |\varphi(y) - A|$.

By existence of the double limit, when $|x-a|, |y-b| < \delta$ we have $|f(x,y) - A| < \epsilon$.

Thus, for all $y$ such that $|y-b| < \delta$,

$$|\varphi(y) - A|= \lim_{x \to a}|f(x,y) - A| \leqslant \epsilon,$$

(Try to justify the RHS inequality on your own).

Therefore,

$$\lim_{y \to b}\lim_{x \to a}f(x,y) = \lim_{y \to b}\varphi(y) = A$$

RRL
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  • why RHS inequality has an equal？ – maplgebra May 14 '21 at 00:59
  • @maplgebra: Since $G(x) = |f(x,y)-A| < \epsilon$ for all $x$ close to a we can't have $\lim_{x \to a} G(x) > \epsilon$ since $(\epsilon, \infty)$ is an open neighborhood. (If the limit were in this neighborhood then there would be values of $x$ close to $a$ for which $G(x) \in (\epsilon,\infty)$ -- impossible.) A limit however can attain the upper bound. For example $1/n < 1$ for all $n\in \mathbb{N}$ (strict inequality), but $\lim_{n \to \infty} 1/n = 1$. – RRL May 14 '21 at 01:19
  • $\lim\limits_{n \rightarrow \infty } \frac1n$ should be 0 ? – maplgebra May 14 '21 at 02:06
  • if let it be $\lim\limits_{n \rightarrow 0}{\frac{1}{n+1}}$ can make sense – maplgebra May 14 '21 at 02:14
  • I meant $1-1/n <1$ and $1-1/n\to 1$ as $n \to \infty$. – RRL May 14 '21 at 02:19
  • thanks for your proof and response! – maplgebra May 14 '21 at 02:35