# How to make a function continuous

ammsa

## Homework Statement

consider the function f(x) = (4 - x) / (2 - $$\sqrt{x}$$). define a new function g(x) = f(x) for all x except 4 and such g(x) is continuous at 4.

## The Attempt at a Solution

i got the limit of the f(x) when x approches 4 and i got 4 as the final answer.
here's how i did it,

1) (4-x) / (2 - $$\sqrt{x}$$ )
2) (2 + $$\sqrt{x}$$) (2 - $$\sqrt{x}$$) / (2 - $$\sqrt{x}$$)
3) (2 + $$\sqrt{x}$$)
4) when we plug in 4 in the equation (2 + $$\sqrt{x}$$), we get
(2 + $$\sqrt{4}$$) = 4

so, g(x) = f(x) when x $$\neq$$ 4
and = (2 + $$\sqrt{x}$$) when x = 4

now my question is is my way of approching this problem correct, and is my answer correct?

Last edited:

## Answers and Replies

Gold Member
I'd say your way of approaching the problem is correct, since it got the correct answer.

Homework Helper
hi ammsa!

(have a square-root: √ )

yup, your steps 1) to 4) are perfect

(but in the last line it would be simpler to say "g(x) = 4 when x = 4")

Homework Helper
By the way, we would say that the function
$$f(x)= \frac{x- 4}{x- \sqrt{2}}$$
has a 'removable' discontinuity at $x= \sqrt{2}$. That is why it was possible to make it continuous.

It is also correct that the $g(x)= x+ \sqrt{2}$ is NOT the same function as f.

Some people make the mistake of writing
$$\frac{x-4}{x-\sqrt{2}}= \frac{(x-\sqrt{2})(x+\sqrt{2})}{x- \sqrt{2}}= x+ \sqrt{2}$$
but that last inequality is only true for $x\ne \sqrt{2}$and not for all x.