Question

If \(\lim_{x\rightarrow 0}\frac{(x(1+acos(x))-bsin(x))}{x^3}=1\), then find the value of a and b.

(a) 2.5, -1.5

(b) -2.5, -1.5

(c) -2.5, 1.5

(d) 2.5, 1.5

I had been asked this question in my homework.

This key question is from Limits and Derivatives of Several Variables topic in portion Partial Differentiation of Engineering Mathematics

Answer 1

The correct choice is (b) -2.5, -1.5

Easy explanation: \(\lim_{x\rightarrow 0}\frac{(x(1+acos(x))-bsin(x))}{x^3}=1\)

Expanding terms of cos(x) and sin(x) and rearranging we get,

\(\lim_{x\rightarrow 0}\frac{(1+a-b)x+(\frac{b}{6}-\frac{a}{2})x^3+(\frac{a}{24}-\frac{b}{120})x^5+….}{x^3}=1\)

Since, given limit is finite, hence coefficients of powers of x should be zero and x^3 should be 1

⇒ 1 + a – b=0

⇒ ^b⁄6 – ^a⁄2 = 1

⇒ Solving the above two equations we get, a = -2.5, b = -1.5.