Question

The values of $a$ and $b$ if $f$ is continuous at $x=0$ where $f\left(0\right)=3$, $f\left(x\right)={(1+\frac{ax+b{x}^3}{x^2})}^{1/x},x>0$

• A. $a=0,b=\log 3$
• B. $a=1,b=\log 2$
• C. $a=2,b=\log 3$
• D. $a=0,b=\log 2$

Answer 1

Answer: (A)

$a=0,b=\log 3$

At $x=0,f\left(x\right)$ is of the form $1^{\infty }$

since the function is continuous ${lim}_{x\to 0}f\left(x\right)=f\left(0\right)$

so $\Rightarrow {lim}_{x\to 0}f\left(x\right)=e^{g\left(x\right)\left[f\right(x)-1]}$

$\Rightarrow {lim}_{x\to 0}f\left(x\right)$= $e^{[1+\frac{ax+b{x}^3}{x^2}-1]1/x}$

$\Rightarrow {lim}_{x\to 0}f\left(x\right)$= $e^{\left[\frac{ax+b{x}^3}{x^3}\right]}$

$\Rightarrow {lim}_{x\to 0}f\left(x\right)$= 3

so, $e^{\left[\frac{ax+b{x}^3}{x^3}\right]}$=3

so, $\left[\frac{ax+b{x}^3}{x^3}\right]=\log 3$

so, $a=0$ and $b=\log 3$