Question

Evaluate $\underset{x\to a}{lim}\frac{1}{(a^2-x^2{)}^2}[\frac{a^2+x^2}{ax}-2\sin \left[\frac{a\pi }{2}\right]\sin \left[\frac{\pi x}{2}\right]]$ where a is an odd integer

• A. $\frac{{\pi }^2{a}^2+4}{16a^4}$
• B. $\frac{{\pi }^2{a}^2-4}{16a^4}$
• C. $\frac{{\pi }^2{a}^2-4}{16a^2}$
• D. $\frac{{\pi }^2{a}^2+4}{16a^2}$

Answer 1

The correct option is A $\frac{{\pi }^2{a}^2+4}{16a^4}$

$\underset{x\to a}{lim}\frac{1}{(a^2-x^2{)}^2}[\frac{a^2+x^2}{ax}-2\sin \left[\frac{a\pi }{2}\right]\sin \left[\frac{\pi x}{2}\right]]$
$=\underset{h\to 0}{lim}\frac{1}{(a^2-a^2-h^2-2ah{)}^2}[\frac{a^2+a^2+h^2+2ah}{a^2+ah}-2\sin \left[\frac{a\pi }{2}\right]\sin \left[\right(a+h\left)\frac{\pi }{2}\right]]$
$=\underset{h\to 0}{lim}\frac{1}{h^2(h+2a{)}^2}[\frac{2a^2+h^2+2ah}{a^2+ah}-[\cos \frac{h\pi }{2}-\cos [a\pi +\frac{h\pi }{2}]]]$
$=\underset{h\to 0}{lim}\frac{1}{h^2(h+2a{)}^2}[\frac{2a^2+h^2+2ah}{a^2+ah}-2\cos \frac{\pi h}{2}]$
[$\because$ a is odd integer]
$=\underset{h\to 0}{lim}\frac{1}{h^2(h+2a{)}^2}[\frac{h^2}{a^2+ah}+2-2\cos \frac{\pi h}{2}]$
$=\underset{h\to 0}{lim}\frac{1}{4a^2}[2\left[\frac{1-\cos \left(\pi /2\right)h}{h^2}\right]+\frac{1}{a^2}]=\frac{{\pi }^2{a}^2+4}{16a^4}$