Question

$f\left(x\right)=\frac{a+b^{\frac{1}{x}}}{c+d^{\frac{1}{x}}},b>1,d>1,c\ne 0,$$f\left(0\right)=1$ is left continuous at $x=0$, then

• A. $a=0$
• B. $a=2c$
• C. $a=c$
• D. $a\ne c$

Answer 1

The correct option is C $a=c$

$f\left(x\right)=\frac{a+b^{1/x}}{c+d^{1/x}},b>1,d>1,f\left(0\right)=1$

$f\left(x\right)$ is left continuous at $x=0$, so $\underset{x\to 0^{-}}{lim}f\left(x\right)=f\left(0\right)$

$x=0-h$

$\underset{x\to 0^{-}}{lim}\frac{a+b^{-\frac{1}{x}}}{x+d^{-1/x}}=1$

$b>1,d>1$

$-\frac{1}{x}\to -\infty \Rightarrow b^{-1/x}\to 0,d^{-1/x}\to 0$

$\underset{x\to 0}{lim}\frac{a}{c}=1$

$a=c$