Question

Find how many rounds of differentiation are required to have finite limit for \(lt_{x\rightarrow 0}\frac{cos(ax)+cos(bx)-2cos(cx)}{cos(ax)+2cos(bx)-3cos(cx)}\) given that  a ≠ b ≠ c

(a) 3

(b) 0

(c) 2

(d) 4

I got this question in examination.

This interesting question is from Indeterminate Forms topic in division Differential Calculus of Engineering Mathematics

Answer 1

Right choice is (c) 2

Easy explanation: Applying L hospitals rule

\(=lt_{x\rightarrow 0}\frac{a.cos(ax)+b.cos(bx)-2c.cos(cx)}{a.cos(ax)+2b.cos(bx)-3c.cos(cx)}=lt_{x\rightarrow 0}\frac{a+b-2c}{a+2b-3c}\)

Assume now that

a + b + 2c = 0 and a + 2b – 3c = 0

We must have

a = c = b but given a ≠ c ≠ b

 Thus, our assumption is false and a finite limit exists after first round of differentiation.

Hence, 2 is the right answer.