Calculus proof - Possible application of a known proof

Question

Let $a_n < c < b_n$ and $a_n, b_n → c$ as $n → ∞ $.

Suppose $ f : \mathbb{R} → \mathbb{R} $ is continuous.

Show that (Possibly using similar proof to FTC...?) $$ \lim_{n \to\infty }\frac{1}{{b_n - a_n}}\int_{{\,a_n}}^{{\,b_n}}{{f\left(x\right)\,dx}} = f(c).$$

Answer 1

Note that $$\min_{a_n\leq x\leq b_n}f(x)\leq\frac1{b_n-a_n}\int_{a_n}^{b_n}f(x)\ \mbox{d}x\leq\max_{a_n\leq x\leq b_n}f(x).$$ Show that the left and right converge to $f(c)$ using the fact that $f$ is continuous and then use the squeeze theorem.

Answer 2

If you already acknowledge Integral Mean Value, then $\dfrac{1}{b_{n}-a_{n}}\displaystyle\int_{a_{n}}^{b_{n}}f(x)dx=f(\eta_{n})$ for some $\eta_{n}$ in between $a_{n}$ and $b_{n}$. Taking $n\rightarrow\infty$ and squeezing $a_{n},b_{n}\rightarrow c$ to get $\eta_{n}\rightarrow c$ and continuity of $f$ implies $f(\eta_{n})\rightarrow f(c)$.

Edit:

By @SmileyCraft answer, since $f$ is continuous, the maxima and minima are attained: \begin{align*} f(\alpha_{n})\leq\int_{a_{n}}^{b_{n}}f(x)dx\leq f(\beta_{n}) \end{align*} for some $a_{n}\leq\alpha_{n},\beta_{n}\leq b_{n}$. Now use Intermediate Value Theorem to get some $a_{n}\leq\eta_{n}\leq b_{n}$ such that \begin{align*} \dfrac{1}{b_{n}-a_{n}}\int_{a_{n}}^{b_{n}}f(x)dx=f(\eta_{n}). \end{align*}

Answer 3

Let $\varepsilon > 0$ be given. Since $f$ is continuous at $c$, there exists $\delta > 0$ so that for all $x$ with $|x-c| < \delta$ we have $|f(x)-f(c)| < \varepsilon$.

Now also we have $a_n \to c$ and $b_n \to c$, so there exists $N \in \mathbb N$ so that for all $n \ge N$ we gave $c-a_n<\delta$ and $b_n - c< \delta$.

Fix $n \ge N$. Then for all $x \in [a_n,b_n]$ we have $|x-c| < \delta$. [Indeed, if $x<c$ then $a_n \le x \le c$ so $|x-c| < \delta$; and similarly if $x>c$. Finally, it is also true when $x=c$.] Thus, for all $x \in [a_n,b_n]$ we have $|f(x)-f(c)| < \varepsilon$. Compute $$ \frac{1}{b_n-a_n}\int_{a_n}^{b_n} f(x)\;dx - f(c) = \frac{1}{b_n-a_n}\int_{a_n}^{b_n} \big(f(x)-f(c)\big)\;dx < \frac{1}{b_n-a_n}\int_{a_n}^{b_n} \varepsilon\;dx = \varepsilon $$ and similarly $$ \frac{1}{b_n-a_n}\int_{a_n}^{b_n} f(x)\;dx - f(c) > -\varepsilon $$

Finally, then: for any $\varepsilon > 0$ there is $N \in \mathbb N$ so that if $n > N$ we have $$ \left|\frac{1}{b_n-a_n}\int_{a_n}^{b_n} f(x)\;dx - f(c)\right| < \varepsilon . $$

Note: we do not require continuity of $f$ everywhere, merely at the one point $c$.