Question

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