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 $$x>1$$ 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 $$\sqrt{1-x}$$