Q.244. Multiple Choice Question on Limits

Question 244.
Given that lim (x → 1) (ax^4 + bx^3 + 1) / [(x-1) sin (πx)] exists and is a finite number, find its value.
(A) 3/π,          (B) - 6/π,          (C) 2/π,          (D) - 4/π,         
(E) not determinable due to insufficient information about a and b.

Answer 244.
If the limit exists, as the denominator becomes zero for x = 1, it should be so for the numerator also.
=> a + b + 1 = 0 ... (1)
Using L'Hospital's theorem,
limit
= lim (x → 1) (4ax^3 + 3bx^2) / [π(x-1) cos (πx) + sin(πx)]
Again, plugging x = 1 in the denominator makes it zero, so should it in the numerator also
=> 4a + 3b = 0 ... (2),
Solving equations (1)and (2), a = 3 and b = - 4

Applying L'Hospital's theorem again,
limit
= lim (x → 1) (12ax^2 + 6bx) / [- π^2(x-1) sin (πx) + π cos(πx) - πcos(πx)]
= (12a + 6b) / (- 2π)
= (6a + 3b) / (- π)
Plugging a = 3 and b = - 4
limit = - 6/π.

Answer: (B) - 6/π.

Link to YA!