# Show that f(x)=2a+3b is continous, where a and b are constants?

Mar 29, 2018

See explanation below

#### Explanation:

A function is called continuous in ${x}_{0}$ if

$f \left({x}_{0}\right)$ exists and is finite

${\lim}_{x \to {x}_{0}} f \left(x\right)$ exists and finally

${\lim}_{x \to {x}_{0}} f \left(x\right) = f \left({x}_{0}\right)$

In our case $f \left(x\right) = 2 a + 3 b$ with $a$ and $b$ constants. So the function doesn't depend of x and for this reason is constant for every value of x, and his value is $f \left(x\right) = 2 a + 3 b$ whatever be a and b.

For this reason

$f \left({x}_{0}\right) = 2 a + 3 b$ exists for every ${x}_{0}$ and is finite
${\lim}_{x \to {x}_{0}} f \left(x\right) = 2 a + 3 b$. This proof that $f \left(x\right)$ is continous in all $x \in \mathbb{R}$