Question

using the sequential definition of continuity, prove that the function f :r to r , defined by f (x) = 3x^2+7, for all x belong to r, is continuous

Answer 1

Sequential definition of continuity states if f is continuous at a if and only if $f(x_n) \to f(a)$ for all sequences $x_n \to a$ .

Given function is $f (x) = 3x^2+7$ .

Let ${<x_n>}$ be any sequence convergence to $a$.

Now, $f(x_n) - f(a) = (3x_n^2+7) - (3a^2+7) = 3(x_n^2-a^2)$

So, $|f(x_n)-f(a)|= |3(x_n+a)(x_n-a)| \leq 3|x_n+a||x_n-a|$ .

Now, we have $x_n\to a$ so ${<x_n>}$ is bounded sequence $\implies |x_n|\leq k \hspace{0.05 in} \forall \hspace{0.05 in} n \geq m$ for some finite m.

$⟹∣x_n+a∣≤k+a$ for all $n≥m$ .

So, $|f(x_n)-f(a)| \leq 3|x_n+a||x_n-a| \leq 3(k+a) |x_n-a|$ for all $n\geq m$.

Thus, as $x_n \to a$, $f(x_n​)→f(a).$

So, by Sequential definition of continuity, f is continuous at every real number.