Calculate Limit: $\frac{\sqrt{z}-i+\sqrt{z+1}}{\sqrt{z^2-1}}$

[jjr]

Homework Statement

Calculate the following limit if it exists
$lim_{z\to -1}\frac{\sqrt{z}-i+\sqrt{z+1}}{\sqrt{z^2-1}}$
the branch of root is chosen to that $\sqrt{-1}=i$

Homework Equations

3. The Attempt at a Solution
[/B]
By inserting $z=-1$ directly, I get a $\frac{0}{0}$ expression.

I don't see how the expression can be simplified before inserting the limit, that seems out of the question.
I have a few theorems involving points at infinity, but none of them apply here.
I considered making the denominator real by multiplying by its complex conjugate, but the numerator will still be 0 when multiplied by any number.

The answer is supposed to be $1/(i\sqrt{2})$.

Not looking for a solution or quick fix, would be very happy if someone could help me the first step of the way.J

 

[FaroukYasser]

Homework Statement

Calculate the following limit if it exists
$lim_{z\to -1}\frac{\sqrt{z}-i+\sqrt{z+1}}{\sqrt{z^2-1}}$
the branch of root is chosen to that $\sqrt{-1}=i$

Homework Equations

3. The Attempt at a Solution
[/B]
By inserting $z=-1$ directly, I get a $\frac{0}{0}$ expression.

I don't see how the expression can be simplified before inserting the limit, that seems out of the question.
I have a few theorems involving points at infinity, but none of them apply here.
I considered making the denominator real by multiplying by its complex conjugate, but the numerator will still be 0 when multiplied by any number.

The answer is supposed to be $1/(i\sqrt{2})$.

Not looking for a solution or quick fix, would be very happy if someone could help me the first step of the way.J

Hint: Try multiplying the expression by: $\frac { \sqrt { z + i } }{ \sqrt { z + i } }$

Moderator note: I revised the expression above to convey what FaroukYasser meant, but did not write.

 

[jjr]

Thank you, got it now!

$\lim_{i\to -1} \frac{\sqrt{z}-1+\sqrt{z+1}}{\sqrt{z^2-1}}$

$\to \lim_{i\to -1} \frac{\sqrt{z}-1+\sqrt{z+1}}{\sqrt{(z+1)(z-1)}}$

$\to \lim_{i\to -1} \frac{1}{\sqrt{z-1}} = \frac{1}{i\sqrt{2}}$

 

[Ray Vickson]

Homework Statement

Calculate the following limit if it exists
$lim_{z\to -1}\frac{\sqrt{z}-i+\sqrt{z+1}}{\sqrt{z^2-1}}$
the branch of root is chosen to that $\sqrt{-1}=i$

Homework Equations

3. The Attempt at a Solution
[/B]
By inserting $z=-1$ directly, I get a $\frac{0}{0}$ expression.

I don't see how the expression can be simplified before inserting the limit, that seems out of the question.
I have a few theorems involving points at infinity, but none of them apply here.
I considered making the denominator real by multiplying by its complex conjugate, but the numerator will still be 0 when multiplied by any number.

The answer is supposed to be $1/(i\sqrt{2})$.

Not looking for a solution or quick fix, would be very happy if someone could help me the first step of the way.J

Is your numerator equal to $\sqrt{z} \: - i \: + \sqrt{z+1}$, or is it $\sqrt{z-i} + \sqrt{z+1}$ (or maybe even $\sqrt{z-i} + \sqrt{z+i}$ )? You wrote the first.

 

[jjr]

The first