# If $\lim_{n\to\infty}a_n=L$ and $f$ is continuous at $L$, then $\lim_{n\to\infty}f(a_n)=f(L)$. Is this true?

Theorem: Prove or disprove the following:

If $\lim_{n\to\infty}a_n=L$ and the function is continuous at $L$, then $\lim_{n\to\infty}f(a_n)=f(L)$.

I'm trying to study for my upcoming exam and I'm struggling to prove the above theorem. I believe you need to use the $\epsilon$-$\delta$ proof method to prove it, but I cannot find a way to do it. I'm not very comfortable with those kinds of proofs just yet. A proof with explanation or some guidance would be greatly appreciated.

• thank you magicman for the edit – user229268 Apr 6 '15 at 21:47
• You're welcome. :) – Daniel W. Farlow Apr 6 '15 at 21:48
• Start by writing down 1) the definition of continuity of $f$ at $x=L$ 2) the definition of what $\lim a_n = L$ means. Then its just a matter of combinding the two to get the proof. – Winther Apr 6 '15 at 21:49

Recall what it means for $a_{n}$ to converge to a limit $L$:

Definition. We say a sequence $a_{n}$ converges to $L$ (in symbol, $a_{n} \to L$, or $\lim \limits_{n \to \infty} a_{n} = L$) if for every $\epsilon > 0$, there is some $N$ such that $n \geq N$ implies $|a_{n} - L| < \epsilon$.

In ordinary words, we say the sequence converges to $L$ if for every positive distance $\epsilon$ around $L$, we can find some point $N$ in the sequence such that for all terms after that point, they are within $\epsilon$-distance of $L$.

Now recall what it means for a function $f : \Bbb R \to \Bbb R$ to be continuous:

Definition. We say $f : \Bbb R \to \Bbb R$ is continuous at $x \in \Bbb R$ if for every $\epsilon > 0$ there is some $\delta > 0$ such that $|x - a| < \delta$ implies $|f(x) - f(a) | < \epsilon$.

In ordinary words, we say $f$ is continuous at $x$ if for each $\epsilon$-distance around $f(x)$, we can find some $\delta$-distance around $x$ such that all points within that $\delta$-distance of $x$ map to within the $\epsilon$-distance of $f(x)$. In some sense, we are saying no matter how close to $f(x)$ we choose, we can find a radius around $x$ such that all points in that radius will be as close as we chose to $f(x)$.

Note that we say $f$ is continuous if it is continuous at each point in its domain.

Now, the problem we need to prove is:

If $f: \Bbb R \to \Bbb R$ is a continuous function and $\{ a_{n} \} \subseteq \Bbb R$ is a sequence such that $a_{n} \to L$, then $f(a_{n}) \to f(L)$.

Proof. All we need to show is that the sequence $\{ f(a_{n} ) \}$ of real numbers converges to the real number $f(L)$. If you look at the definition above, that means we need to show for every $\epsilon > 0$, there is some $N$ such that $n \geq N$ implies $|f(a_{n}) - f(L)| < \epsilon$.

Now, we have to use each assumption we've made. One of the assumptions is that $f$ is continuous. Since $f(L)$ is a fixed number (because $L$ is a fixed number), it looks like we will want to use the fact that $f$ is continuous at $L$ in some way (since it is continuous everywhere). Let's write down what it means for $f$ to be continuous at $L$:

It means for every $\epsilon > 0$ there is some $\delta > 0$ such that $|L - a| < \delta$ implies $|f(L) - f(a)| < \epsilon$.

So, let's prove $f(a_{n})$ converges to $f(L)$: Let $\epsilon > 0$ be arbitrary. We know $f$ is continuous at $L$, so there is some $\delta > 0$ such that $|L - a| < \delta$ implies $|f(L) - f(a)| < \epsilon$. But we also know that $a_{n} \to L$, which means in the $\delta$-distance around $L$ that we just found via continuity, there is some $N$ such that $n \geq N$ implies $|a_{n} - L| < \delta$. But by continuity, since for all $n \geq N$, $|a_{n} - L| < \delta$, then $|f(a_{n}) - f(L)| < \epsilon$. But $\epsilon$ was arbitrary, and this is true for all $n \geq N$, and so the last statement was exactly what it means for $\{ f(a_{n}) \}$ to converge to $f(L)$. Thus, $f(a_{n}) \to f(L)$, as desired.

Let $\epsilon > 0$,

1. As $f$ is continuous at $L$, there exist $\delta$ such that $|x-L|< \delta \Rightarrow |f(x)-f(L)| < \epsilon$

2. As $a_n \to L$, there exist a rank $n_0$ such that $\forall n > n_0, |a_n-L| < \delta$

1+2 gives you that there exist a rank $n_0$ such that $\forall n > n_0, |f(a_n)-f(L)| < \epsilon$

And you have the convergence

First, of all, you need to assume that $f$ is defined in a neighborhood of $L$, otherwise, it may not even make sense! For example, consider the function $f:\{0\} \to \mathbb{R}$ defined by $f(x) = 0$. This function is clearly continuous at $0$ but it does not even make sense to ask what $f(a_n)$ is, as the function is not defined.

Let us assume that there is an open interval $I$, containing $0$, such that $I\subseteq \text{dom}(f)$. Further, we assume that $a_n\in I$ with $a_n\to 0$. With this assumption, the sequence of values, $f(a_n)$ now makes sense.

Let $\varepsilon > 0$. Since $f$ is continuous at $0$ there is an open interval $J\subset I$, so that $f(J) \subseteq (L-\varepsilon,L+\varepsilon)$. But $a_n\to 0$, therefore, by definition of convergence, there is an $N>0$ such that if $n>N$ then $a_n\in J$. From here it follows that $|f(a_n) - L| < \varepsilon$.

• i wonder why you raise the point of neighborhood of $L$. If a function $f$ is continuous at $L$ then it already means that $f$ is defined in a certain neighborhood of $L$. – Paramanand Singh Apr 9 '15 at 3:03

Since $\; f$ is continuous at $L$, then for each $\epsilon > 0$, there exists $\delta > 0$ such that $x \in dom(f)$ and $|x - L| < \delta$ imply $|f(x) - f(L)| < \epsilon$

Since $\lim a_n = L$, there exists an N so that $$n > N \implies |a_n - L| < \delta$$ It follows that $$n > N \implies |f(a_n) - f(L)| < \epsilon$$ This prove that $\lim f(a_n) = f(L)$