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Let $\left(u_n\right)$ be a succession with general term: $$u_n=3+\frac{(-1)^n}{n}$$ and $h$ a real function such that $\lim h\left(u_n\right)=6$.

This is a multiple choice question and I'm pretty sure the right answer is the following:

If $h$ is continuous in $x=3$ then $h(3)=6$.

But why can't we say that $\lim\limits_{x\to 3}h(x)=6$ (this is another option)?

I'm thinking that if we have $n$ even or odd we could define two successions such that $v_n\to 3^-$ and $w_n\to 3^+$ and that would mean $\lim\limits_{x\to 3}h(x)=6$. Am I wrong?

Concept7
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  • The limit exists iff lim $h(a_n)$ exists for every sequence that converges to 3. Knowing that $h$ is continuous at 3 is necessary for us to know that this is true from just one example. –  Sep 22 '19 at 15:23
  • Two successions are not enough. Every conceivable possible uncountably many infinite concessions must all got to six. – fleablood Sep 22 '19 at 15:24

2 Answers2

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Suppose that$$h(x)=\begin{cases}6&\text{ if }x=u_n\text{ for some }n\in\mathbb N\\0&\text{ otherwise.}\end{cases}$$Then you don't have $\lim_{x\to3}h(x)=6$, but it is still true that $\lim_{n\to\infty}h(u_n)=6$.

José Carlos Santos
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The point is this:

$\lim_{n\to\infty}h(x_n)=6$ means that the sequence $(h(x_n))_{n\in\mathbb{N}}$ (with the given sequence $(x_n)_{n\in\mathbb{N}}$) converges to 6.

But $\lim_{x\to 3}h(x)=6$ means, that for every sequence $(x_n)_{n\in\mathbb{N}}\subset \mathbb{R}\setminus \{3\}$ that converges to $3$ it holds that the sequence $(h(x_n))_{n\in\mathbb{N}}$ converges to $6$.

Now obviously, from the first one, you can't follow the secound one.

S. M. Roch
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