Question

If \( \alpha \) is a root of \( x^{2}+a x+1=0 \), then \( \lim _{x \rightarrow 1 / \alpha} \frac{\sin \left(x^{2}+a x+1\right)}{(\alpha x-1)} \) is equal to

Answer 1

 ∵ α is root of x² + ax + 1 = 0

∴ α² + aα + 1 = 0

\(\lim\limits_{x\to 1/α}\frac{sin(x^2+ax+1)}{αx-1}\) (0/0 case)

\(=\lim\limits_{x\to 1/α}\frac{cos(α^2+ax+1)(2x+a)}{\alpha}\) (By using D.L.H. rule)

= \(\frac{cos(1/α^2+a/α+1)(2/α + a)}{α}\)

= \(\frac{cos(1/α^2(α^2+aα+1))(2/α+a)}{α}\) 

= 2/α²  + a/α cos 0 ( ∵α² + aα + 1 = 0)

= 2/α²  + a/α (∵ cos 0 = 1) 

Hence,\(\lim\limits_{x\to 1/α}\frac{sin(x^2+ax+1)}{αx-1}\) = 2/α²  + a/α