In the following function .

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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

  • $\begingroup$ Your left limit is not correct. $\endgroup$ – hamam_Abdallah Dec 22 '16 at 17:02

The limit $x\to \frac{1}{2}_+$ of $f$ formally gives us


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?

  • $\begingroup$ I could not understand what do you mean by common trick $\endgroup$ – user123733 Dec 22 '16 at 16:48
  • $\begingroup$ I mean that this is a technique which is often used in these types of problems. As another example: $\frac{\sqrt{x-2}}{\sqrt{x-1}-1}$ can be manipulated by multiplying by $1$ in a clever way: multiply numerator and denominator by $\sqrt{x-1}+1$ (the conjugate of the denominator). If you do the calculations you will see the denominator will simplify significantly. Oftentimes, this results in the necessary cancellation to compute the limit. $\endgroup$ – Matt Dec 22 '16 at 16:54

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.

  • $\begingroup$ The function you have used is for x<1/2 $\endgroup$ – user123733 Dec 22 '16 at 16:16
  • $\begingroup$ Yes , I am okay with that $\endgroup$ – user123733 Dec 22 '16 at 16:47

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