# How do you use the definition of continuity and the properties of limits to show the function is continuous F(x)= x+sqrt(x-1) on the interval [1, inf)?

Feb 9, 2017

see below

#### Explanation:

For $a > 1$

We need to show that $F \left(a\right) = {\lim}_{x \to a} F \left(x\right)$. So

$F \left(a\right) = a + \sqrt{a - 1}$

${\lim}_{x \to a} \left(x + \sqrt{x - 1}\right) = {\lim}_{x \to a} x + {\lim}_{x \to a} \sqrt{x - 1}$

=lim_(x->a) x + sqrt(lim_(x->a) (x-1)

$= {\lim}_{x \to a} x + \sqrt{{\lim}_{x \to a} x - {\lim}_{x \to a} 1}$

$= a + \sqrt{a - 1}$

Since $F \left(a\right) = {\lim}_{x \to a} F \left(x\right) = a + \sqrt{a - 1}$ therefore

F is continuous at x = a for every a in $\left(1 , \infty\right)$

We also need to show that $F \left(1\right) = {\lim}_{x \to {1}^{+}} F \left(x\right)$

$F \left(1\right) = 1 + \sqrt{1 - 1} = 1 - 0 = 1$

${\lim}_{x \to {1}^{+}} F \left(x\right) = {\lim}_{x \to {1}^{+}} \left(x + \sqrt{x - 1}\right)$

=lim_(x->1^+) x + sqrt(lim_(x->1) (x-1)

$= {\lim}_{x \to {1}^{+}} x + \sqrt{{\lim}_{x \to {1}^{+}} x - {\lim}_{x \to {1}^{+}} 1}$

$= 1 + \sqrt{1 - 1} = 1 + 0 = 1$

Since $F \left(1\right) = {\lim}_{x \to {1}^{+}} F \left(x\right) = 1$ and F is continuous at x = a for

every a in $\left(1 , \infty\right)$ thus F is continuous on $\left[1 , \infty\right)$