Question

If $\underset{x\to 0}{lim}\frac{a\sin x-bx+{cx}^2+x^3}{2x^2\log (1+x)-2x^3+x^4}=I$ and $I$ is finite, then

• A. $a=6$
• B. $b=3$
• C. $c=0$
• D. $I=\frac{3}{40}$

Answer 1

The correct option is A $a=6$

${lim}_{x\to 0}\left\{\frac{asinx-bx+c{x}^2+x^3}{2x^{2}log(1+x)-2x^3+x^4}\right\}$

this is $\frac{0}{0}$ form so applying L-hospital rule, we get

${lim}_{x\to 0}\left\{\frac{acosx-b+2cx+3x^2}{\frac{2x^2}{1+x}+4xlog(1+x)-6x^2+4x^3}\right\}$

this is also $\frac{0}{0}$ form so applying L-hospital rule, we get

${lim}_{x\to 0}\left\{\frac{-asinx+2c+6x}{\frac{2\ast 2x(1+x)-2x^2\left(1\right)}{(1+x{)}^2}+\frac{4x}{(1+x)}+4log(1+x)-12x+12x^2}\right\}$

this is again $\frac{0}{0}$ form so applying L-hospital rule, we get

${lim}_{x\to 0}\left\{\frac{-acosx+6}{\frac{d}{dx}(\frac{2\ast 2x(1+x)-2x^2}{(1+x{)}^2}+\frac{4x}{(1+x)}+4log(1+x)-12x+12x^2)}\right\}$

putting x=0, we get

$=\frac{a-6}{12}$

now the value of a can be founded by equation the above equation equal to 0 i.e,

$=\frac{a-6}{12}=0$

$=a-6=0$

$\Rightarrow a=6.$

Hence, option (a) is correct.