Question

If $\underset{x\to 1}{lim}\frac{a{x}^2+bx+c}{(x-1{)}^2}=2$. FInd $(a,b,c)$.

• A. $(2,4,2)$
• B. $(2,-4,-2)$
• C. $(2,-4,2)$
• D. $(-2,4,2)$

Answer 1

The correct option is C $(2,-4,2)$

If the limit is existing, then it must be $0/0$ form because denominator is zero at $x=1$

$\therefore a+b+c=0\dots \dots \left(1\right)$

Now, using L'hopital rule, we get

${lim}_{x\to 1}\frac{2ax+b}{2(x-1)}=2$

$\because (x-1)=0$ at $x=1$

$\therefore 2a+b=0\dots \dots \left(2\right)$

$0/0$ form, again applying L'hopital rule

${lim}_{x\to 1}\frac{2a}{2}=2⟹a=2$

From eqn (2) and (1)

$b=-4$ and $c=2$

$\therefore (a,b,c)=(2,-4,2)$