Find the value of p for the functionto be continous In the following function .

We have to determine the value of p , if possible, so that the function is
continuous at $x = 1/2$.
I tried and found the LHS of function at $x=1/2$ that is $(-1/4)$. 
But how could we comment on RHS
 A: The limit $x\to \frac{1}{2}_+$ of $f$ formally gives us
$$\frac{\sqrt{2\frac{1}{2}-1}}{\sqrt{4+\sqrt{2\frac{1}{2}-1}}-2}=\frac{0}{0}.$$
This is indeterminate form, and so we can either use L'Hopital's rule (however, given the context, I will avoid this), or we can compute this limit by hand by the common trick of multiplying by the conjugate:
$$\lim_{x\to \frac{1}{2}_+} \frac{\sqrt{2x-1}}{\sqrt{4+\sqrt{2x-1}}-2}= \lim_{x\to\frac{1}{2}_+} \frac{\sqrt{2x-1}(\sqrt{4+\sqrt{2x-1}}+2)}{\sqrt{2x-1}}=\lim_{x\to\frac{1}{2}_+} \sqrt{4+\sqrt{2x-1}}+2=4.$$
Is this sufficient to help you finish the problem?
A: 
Left limit

Put $t=x-\frac{1}{2}$.
then for $x>\frac{1}{2}$
$$f(x)=\frac{1-\cos(\pi t)  }{ 1-\cos(2\pi t) }\sim \frac {\frac{\pi^2 t^2 }{2}} {\frac{4\pi^2 t^2  }{ 2  }   }$$
the left limit is $\frac{1}{4}$.

Right limit

put $\sqrt{2x-1}=4u$, then
$$f(x)=\frac{4u}{2(\sqrt{1+u}-1 )}\sim \frac{4u}{2\frac{u}{2}}$$ which goes to $4$ when $u $ goes to $0^+$.
So we have
$$\frac{1}{4}=\lim_{x\to (\frac{1}{2})^-} f(x)\neq \lim_{x\to (\frac{1}{2})^+}f(x)=4$$
$p$ doesn't exist.
