continuity

Apr 2010
17
0
Can somebody please help me with this? the algebra is making me go crazy.
Let f: [0,1]\(\displaystyle \rightarrow\)R be defined as:

f(x) = 1/\(\displaystyle \sqrt{x}\) - \(\displaystyle \sqrt{((1+x)/x)}\) if x is not equal to 0.

I need to prove that f is continuous on [0,1]. What I want to do is pick a point in [0,1] and show that is is continuous at that point. So i can pick p to be .5. Then f(x)-f(.5) is less than epsilon but the algebra here i am having a hard time. Can someone please help me. Thank you .
 
Oct 2009
4,261
1,836
Can somebody please help me with this? the algebra is making me go crazy.
Let f: [0,1]\(\displaystyle \rightarrow\)R be defined as:

f(x) = 1/\(\displaystyle \sqrt{x}\) - \(\displaystyle \sqrt{((1+x)/x)}\) if x is not equal to 0.

I need to prove that f is continuous on [0,1]. What I want to do is pick a point in [0,1] and show that is is continuous at that point. So i can pick p to be .5. Then f(x)-f(.5) is less than epsilon but the algebra here i am having a hard time. Can someone please help me. Thank you .


\(\displaystyle \frac{1}{\sqrt{x}}-\sqrt{\frac{1+x}{x}}=\frac{1}{\sqrt{x}}-\frac{\sqrt{1+x}}{\sqrt{x}}\) \(\displaystyle =\frac{1-\sqrt{1+x}}{\sqrt{x}}\)

Now, this is a quotient of functions so it is continuous at any point where both the numerator and the denominator are continuous and the latter isn't zero, so

the only problem is at \(\displaystyle x=0\) . Since the function isn't defined at this point, I suppose you meant to show that this point is a removable discontinuity and for that

you need to show the limit of the function exists and is finite there:

\(\displaystyle \lim_{x\to 0}\frac{1-\sqrt{1+x}}{\sqrt{x}}\) , and it's easy to show this limit is zero (with L'Hospital, for example).

Tonio