Question

How do you prove that `h(x) = sqrt(2x - 3)` is continuous as x=2?

Answer 1

A function `f(x)` is said to be continuous at `a` if `lim_(xrarra)f(x)=f(a)`

So in this case, our `f(x)` is `sqrt(2x-3)` and `a` is `2`. Compute the limit, then verify that it equals `h(2)`:

`color(white)=lim_(xrarr2)sqrt(2x-3)`

`=sqrt(2(2)-3)`

`=sqrt(4-3)`

`=sqrt1`

`=1`

Here's `h(2)`:

`h(2)=sqrt(2(2)-3)`

`color(white)(h(2))=sqrt(4-3)`

`color(white)(h(2))=sqrt1`

`color(white)(h(2))=1`

Since the limit and the function are equal, the function `h(x)` is continuous at `x=2`. Hope this helped!