Question

lim x=pi/2 sin ax+ bx / x_pi/2

Answer 1

a cos(aπ/2) + b

Answer 2

The limit is of the form $\lim_{x \to \pi/2} \frac{\sin(ax) + bx}{x - \pi/2}$. For this limit to be finite, it must be of the $\frac{0}{0}$ indeterminate form. This implies that the numerator must be zero at $x = \pi/2$, i.e., $\sin(a\pi/2) + b\pi/2 = 0$. Assuming this condition holds, L'Hopital's Rule can be applied. Differentiating the numerator gives $a\cos(ax) + b$, and differentiating the denominator gives $1$. Substituting $x = \pi/2$ into the ratio of derivatives yields $a\cos(a\pi/2) + b$.

The limit is $a \cos\left(\frac{a\pi}{2}\right) + b$, provided that $\sin\left(\frac{a\pi}{2}\right) + \frac{b\pi}{2} = 0$.