Is the Function f(x)= (x+1)/(1-√(1-x)) Derivable at x0=1⁻?

[mohlam12]

Hey everyone
I was going through some problems and I came across this one, it said to see if the fuction below is derivable in the point x_{0}=1^{-}
f(x)=\frac{x+1}{1-\sqrt{1-x}}

so I did this as usual,
limit of \frac{f(x)-f(1)}{x-1} when x -> 1, and x<1 is equal to the limit of:

\frac{x-1+2\sqrt{1-x}}{(x-1)(1-\sqrt{1-x}}but I really don't know what to do after! I tried to multiply the top and bottom with "x+2sqrt(1-x)" but with no satisfying results.

Any help would be appreciated!

 

[J77]

What do you mean by "derivable"?

Obviously, for x&gt;1 you get complex solutions, and f(x)\rightarrow\pm\infty as x\rightarrow0 (- from left, + from right)...

I wrote it as:

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

 

[mohlam12]

Maybe I meant differentiable... sorry!
what I need to find the limit of is this function f(x)=\frac{x-1+2\sqrt{1-x}}{(x-1)(1-\sqrt{1-x})} as x\rightarrow1^{-}
Thank you

 

[VietDao29]

Wait, can you copy exactly what the problem says please. I dunno, but f(1) is undefined, so the function is not differentiable at x = 1.

 

[J77]

Wait, can you copy exactly what the problem says please. I dunno, but f(1) is undefined, so the function is not differentiable at x = 1.

Why undefined?

Because \sqrt{0}?

As you approach from the left, it tends to 2 tho'...

(and the derivative goes to \infty?)

 

[Curious3141]

Maybe I meant differentiable... sorry!
what I need to find the limit of is this function f(x)=\frac{x-1+2\sqrt{1-x}}{(x-1)(1-\sqrt{1-x})} as x\rightarrow1^{-}
Thank you

The limit of that expression is easy enough to find. Just substitute u = \sqrt{1-x}

But I don't know what your question is asking, or, frankly, what you're trying to accomplish on the whole. Post the question verbatim, please.

 

[Emieno]

f(x)=\frac{x-1}{(x-1)(1-\sqrt{1-x})}+\frac{2}{1-x-\sqrt{1-x}}
the first one's lim=1 and second one's lim=0

 

[VietDao29]

Why undefined?

Because \sqrt{0}?

As you approach from the left, it tends to 2 tho'...

(and the derivative goes to \infty?)

Ack, ack, doing maths late is never good...
Okay, for mohlam12's question, have you considered factoring the \sqrt{1 - x} out in the numerator?
\lim_{x \rightarrow 1 ^ -} \frac{x - 1 + 2 \sqrt{1 - x}}{(x - 1) (1 - \sqrt{1 - x})} = \lim_{x \rightarrow 1 ^ -} \frac{\sqrt{1 - x} (\sqrt{1 - x} - 2)}{(1 - x) (1 - \sqrt{1 - x})} = ...
You can go from here, right? :)
Sorry for such confusion... My bad

 

[mohlam12]

Yup, I can go from here, thanks. I just didn't have that idea to factor with \sqrt{1-x}