Question

If $\underset{x\to 0}{\lim }\frac{\left(1+a^3\right)+8e^{1/x}}{1+\left(1-b^3\right)e^{1/x}}=2$, then

• A. $a=1,b=(-3{)}^{1/3}$
• B. $a=1,b=3^{1/3}$
• C. $a=-1,b=-(3{)}^{1/3}$
• D. None of these

Answer 1

Answer: (A)

We have $2=\underset{x\to 0}{\lim }\frac{\left(1+a^3\right)+8e^{1/x}}{1+\left(1-b^3\right)e^{1/x}}(\frac{\infty }{\infty }$ form $)$
$\Rightarrow 2=\underset{x\to 0}{\lim }\frac{0+8e^{1/x}\left(-1/x^2\right)}{0+\left(1-b^3\right)e^{1/x}\left(-1/x^2\right)}$
(Using L’Hospital’s Rule)
$\Rightarrow 1-b^3=4$
$\Rightarrow b^3=-3$
$\Rightarrow b=(-3{)}^{1/3}$
$\therefore$ From Eq. (1), $2=\underset{x\to 0}{\lim }\frac{\left(1+a^3\right)+8e^{1/x}}{1+4e^{1/x}}$
$\Rightarrow 1+a^3=2$ i.e., $a=1$
Hence $a=1$ and $b=(-3{)}^{1/3}$.