Question

\(\lim_{x\rightarrow 0}\frac{ax^3+b sin(x)+c cos(x)}{x^5}=1\), then find the value of a, b and c.

(a) 1.37, -4.13, 4.13

(b) 1.37, 4.13, -4.13

(c) -1.37, 4.13, 4.13

(d) 1.37, 4.13, 4.13

This question was posed to me during an interview.

The above asked question is from Limits and Derivatives of Several Variables in section Partial Differentiation of Engineering Mathematics

Answer 1

Right answer is (b) 1.37, 4.13, -4.13

Explanation: \(\lim_{x\rightarrow 0}\frac{ax^3+b sin(x)+c cos(x)}{x^5}=1\)

Now expanding the terms of sin(x) and cos(x) and rearranging in powers of x,x^3 and x^5 and so on,we get

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

Now, coefficient of x and x^3 should be zero and that of x^5 should be 1, then

⇒ B + c = 0

⇒ ^b⁄6 + ^c⁄2 = a

⇒ ^b⁄120 + ^c⁄24 = 1

⇒ By solving these 3 equations, a = 1.37, b = 4.13, c = -4.13.