Question

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

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

Answer 1

Answer: (C)

$b=6,c=0$

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

This is in $\frac{0}{0}$ form, so apply L'Hospital's Rule, we get

$\Rightarrow li{m}_{x\to 0}\frac{acosx-b+2cx+3x^2}{\frac{2x^2}{1+x}+\log (1+x)4x-6x^2+4x^3}$

Denominator is $0$. Therefore, for limit to exist, numerator should be $0$

$\Rightarrow a-b=0\Rightarrow a=b$

This is in $\frac{0}{0}$ form, so again apply L'Hospital's Rule, we get

$\Rightarrow \frac{-asinx+2c+6x}{\frac{(1+x)4x-2x^2}{1+x^2}+\frac{4x}{1+x}+4\log (1+x)-12x+12x^2}$

Denominator is $0$. Therefore, for limit to exist, numerator should be $0$

$\Rightarrow c=0$

This is in $\frac{0}{0}$ form, so again apply L'Hospital's Rule, we get

$\Rightarrow \frac{-acosx+6}{\frac{(1+x{)}^2(4+4x)-2(4x+2x^2\left)\right(1+x)}{(1+x^4)}+\frac{(1+x)4-4}{(1+x{)}^2}+\frac{4}{1+x}-12+24x}$

Denominator is $0$. Therefore, for limit to exist, numerator should be $0$

$\Rightarrow -a+6=0\Rightarrow a=6$

$\Rightarrow a=b=6$