1
$\begingroup$

This function is not continuous at $x=0$. I know that function (in the example) is continuous if $$\lim\limits_{x\to0^-}f(x)=\lim\limits_{x\to0^+}f(x)=f(0)$$ and limits and $f(x_0)$ must be defined. I am getting that $$f(0)=\frac{0}{0}$$ which is not defined. What is the condition for $f(x)$ to be continuous at $x=0$? If the condition exists, how to find if $f(x)$ is differentiable at $x=0$?

Thanks for replies.

$\endgroup$

2 Answers 2

1
$\begingroup$

Hint: Frist find $\displaystyle\lim_{x\to0}f(x)$ (Maybe try to look for a substitution). Then redefine the function for it to be continuous at $x=0$. Also recall that a function is differentiable at a point $x_0$ if and only if $f'(x_0)$ exists.

$\endgroup$
0
$\begingroup$

the limit value is 4, so put f(0)=4 and it will be continuous at x=0

http://www.wolframalpha.com/input/?i=\lim_{x+tends+to+0}\frac{%281-x%29^{-1%2F2}-%281%2Bx%29^{1%2F2}}{%281-\frac{x}{2}%29^{-1%2F2}-%281%2B\frac{x}{2}%29^{1%2F2}}

f'(0)=1 (from the maclaurent expansion given by WOlfram, you can also see its graph in color blue)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .