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)?

1 Answer
Feb 9, 2017

see below

Explanation:

For a>1

We need to show that F(a)=lim_(x->a) F(x). So

F(a)=a+sqrt(a-1)

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

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

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

=a+sqrt(a-1)

Since F(a)=lim_(x->a) F(x)=a+sqrt(a-1) therefore

F is continuous at x = a for every a in (1,oo)

We also need to show that F(1)=lim_(x->1^+) F(x)

F(1)=1+sqrt(1-1)=1-0=1

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

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

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

=1+sqrt(1-1)=1+0=1

Since F(1)=lim_(x->1^+) F(x)=1 and F is continuous at x = a for

every a in (1,oo) thus F is continuous on [1,oo)