Question

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

The correct option is A $a=1,b=(-3{)}^{1/3}$

We have,
$\underset{x\to 0}{lim}\frac{(1+a^3)+8e^{1/x}}{1+(1-b^3)e^{1/x}}=2$ $\left[\frac{\infty }{\infty }form\right]$
$\Rightarrow \underset{x\to 0}{lim}\frac{(1+a^3)e^{-1/x}+8}{e^{-1/x}+(1-b^3)}=2$
$\Rightarrow \frac{0+8}{0+(1-b^3)}=2\Rightarrow 1-b^3=4\Rightarrow b^3=-3\Rightarrow b=(-3{)}^{1/3}$
Again,
$\underset{x\to 0}{lim}\frac{(1+a^3)+8e^{1/x}}{1+(1-b^3)e^{1/x}}=2$
$\Rightarrow \underset{x\to 0}{lim}\frac{(1+a^3)+8e^{1/x}}{1+4e^{1/x}}=2$
$\Rightarrow 1+a^3=2$ [On comparison]
$\Rightarrow a=1$
Therefore, $a=1,b=(-3{)}^{1/3}$
Hence, option 'A' is correct.