Question

If f(x) = x^2 – 3x + 2 and g(x) = x^3 – x^2 + x – 1 than find value of \(\lim_{x\rightarrow 1}\frac{f(x)}{g(x)}\)?

(a) 0.5

(b) 1

(c) -.5

(d) -1

This question was posed to me in unit test.

My question comes from Indeterminate Forms in portion Differential Calculus of Engineering Mathematics

Answer 1

The correct choice is (c) -.5

For explanation: \(\lim_{x\rightarrow 1}\frac{f(x)}{g(x)}=\lim_{x\rightarrow 1}\frac{x^2-3x+2}{x^3-x^2+x-1}\)=(1-3+2)/(1-1+1-1)=0/0

Hence this gives indeterminent forms hence by L’Hospital Rule

\(\lim_{x\rightarrow 1}\frac{f(x)}{g(x)}=\lim_{x\rightarrow 1}\frac{f'(x)}{g'(x)}=\lim_{x\rightarrow 1}\frac{2x-3}{3x^2-2x+1}=\frac{2-3}{3-2+1}=-\frac{1}{2}\)

Hence

\(\lim_{x\rightarrow 1}\frac{f(x)}{g(x)}=-\frac{1}{2}\)