$$\lim_{x\to a}\frac{1}{(a^2-x^2)^2}\cdot\left(\frac{a^2+x^2}{ax}-2\sin\frac{a\pi}{2}\sin\frac{\pi x}2\right)=?$$ if $a$ is an odd integer.

The way I set out is first assuming $a=1$ and seeing if I can spot some pattern. Now, if I rewrite this limit out as:

$$\lim_{x\to 1}\frac{1}{(1-x^2)^2}\cdot\left(\frac{1+x^2}{x}-2\sin\frac{\pi x}2\right)=?$$

Now, you can clearly see that the denominator is of the form $0^4$ (due to the $(1-x^2)^2$), whereas the numerator is, at max, $0^2$ (due to the $x^2$). Since denominator has a higher power of zero than the numerator, I believe that the limit won't exist.

However, my textbook says this limit exists and has a finite value. So, I wish to ask what is the fault in my reasoning.

  • $\begingroup$ $$\lim_{x\to a}\frac{1}{(a^2-x^2)^2}\cdot\left(\frac{a^2+x^2}{ax}-2\sin\frac{ax}{2}\sin\frac{\pi x}2\right)$$ $$=\lim_{x\to a}\dfrac1{(a+x)^2}\left(\lim_{x\to a}\dfrac{(x-a)^2}{ax(x-a)^2}+\lim_{x\to a}\dfrac{2-2\sin\frac{ax}{2}\sin\frac{\pi x}2}{(x-a)^2}\right)$$ $\endgroup$ – lab bhattacharjee Mar 16 '18 at 8:45
  • 1
    $\begingroup$ Maybe $$...-2\sin\frac{\pi a}{2}\sin\frac{\pi x}2$$ $\endgroup$ – user Mar 16 '18 at 9:10
  • $\begingroup$ @gimusi Well spotted, and sorry for the typo. I've fixed it, you're right! Thanks! $\endgroup$ – Gaurang Tandon Mar 16 '18 at 9:11
  • $\begingroup$ The limit exists and is finite. First keep $\frac 1 {(1+x)^{2}}$ away to make things simple and apply L'Hopital's Rule. I think the limit is 1/4 for $a=1$. $\endgroup$ – Kavi Rama Murthy Mar 16 '18 at 9:15

Note that for $a=1$ with $x=1+y$ and $y\to 0$

$$\frac{1}{ (1-x^2)^2 }\cdot\left(\frac{1+x^2}{x}-2\sin\frac{\pi x}2\right) =\frac{1}{ (y^2+2y)^2 }\cdot\left(\frac{y^2+2y+2}{y+1}-2\cos\frac{\pi y}2\right)=...$$

and by Taylor's series

$$\cos \frac{\pi y}2=1-\frac{\pi^2 y^2}8+o(y^2)$$


$$...=\frac{1}{ (y^2+2y)^2 }\cdot\left(\frac{y^2+2y+2}{y+1}-2\cos\frac{\pi y}2\right) =\\=\frac{1}{ y^2(y+2)^2 }\cdot\left(\frac{y^2+2y+2}{y+1}-2+\frac{\pi^2 y^2}4+o(y^2)\right)=\\=\frac{1}{ y^2(y+2)^2 }\cdot\left(\frac{y^2+2y+2-2-2y+\frac{\pi^2 y^2}4+o(y^2)}{y+1}\right)=\\=\frac{1}{ (y+2)^2 }\cdot\left(\frac{1+\frac{\pi^2 }4+o(1)}{y+1}\right)\to\frac14+\frac{\pi^2}{16}$$

For the general case just let $x=az$ with $z\to 1$ and note that

$$\lim_{x\to a}\frac{1}{(a^2-x^2)^2}\cdot\left(\frac{a^2+x^2}{ax}-2\sin\frac{a\pi}{2}\sin\frac{\pi x}2\right)=\\ =\lim_{z\to 1}\frac1{a^4}\frac{1}{(1-z^2)^2}\cdot\left(\frac{1+z^2}{z}-2\sin\frac{a\pi}{2}\sin\frac{\pi a z}2\right)$$

which can be solved with the same approach.

Notably note that for $z=1+y$ and $y\to 0$

$$\frac1{a^4}\frac{1}{ (1-z^2)^2 }\cdot\left(\frac{1+z^2}{z} -2\sin\frac{a\pi}{2}\sin\frac{\pi a z}2 \right) =\frac1{a^4}\frac{1}{ (y^2+2y)^2 }\cdot\left(\frac{y^2+2y+2}{y+1}-2\cos\frac{a \pi y}2\right)=...$$

and by Taylor's series

$$\cos \frac{a\pi y}2=1-\frac{a^2\pi^2 y^2}8+o(y^2)$$


$$...=\frac1{a^4}\frac{1}{ (y^2+2y)^2 }\cdot\left(\frac{y^2+2y+2}{y+1}-2\cos\frac{a\pi y}2\right) =\\=\frac1{a^4}\frac{1}{ y^2(y+2)^2 }\cdot\left(\frac{y^2+2y+2}{y+1}-2+\frac{a^2\pi^2 y^2}4+o(y^2)\right)=\\=\frac1{a^4}\frac{1}{ y^2(y+2)^2 }\cdot\left(\frac{y^2+2y+2-2-2y+\frac{a^2\pi^2 y^2}4+o(y^2)}{y+1}\right)=\\=\frac1{a^4}\frac{1}{ (y+2)^2 }\cdot\left(\frac{1+\frac{a^2\pi^2 }4+o(1)}{y+1}\right)\to\frac1{4a^4}+\frac{\pi^2}{16a^2}$$

  • $\begingroup$ so , if $a$ is an odd integer , limit will be $\frac {1}{4a^4}$. Right ? $\endgroup$ – освящение Mar 17 '18 at 5:44
  • $\begingroup$ @PrithiviRaj the limit in general is $\frac1{4a^4}+\frac{\pi^2}{16a^2}$ $\endgroup$ – user Mar 17 '18 at 8:02

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