Question

Consider the function $f\left(x\right)={\left(\frac{ax+1}{bx+2}\right)}^x$ where $a,b>0$ then $\underset{x\to \infty }{lim}f\left(x\right)$ ____

• A. exists for all values of $a$ and $b$
• B. is zero for $a<b$
• C. is non existent for $a>b$
• D. is $e^{(-\frac{1}{a})}$ or $e^{(-\frac{1}{b})}$ if $a=b$

Answer 1

Answer: (B), (C), (D)

The correct options are
B is zero for $a<b$
C is non existent for $a>b$
D is $e^{(-\frac{1}{a})}$ or $e^{(-\frac{1}{b})}$ if $a=b$

Case 1. $a<b$
$\underset{x\to \infty }{lim}f\left(x\right)=\underset{x\to \infty }{lim}{\left(\frac{ax+1}{bx+2}\right)}^x=\underset{x\to \infty }{lim}{\left(\frac{a+1/x}{b+2/x}\right)}^x=\underset{x\to \infty }{lim}{\left(\frac{a}{b}\right)}^x=0$
Case 2. $a>b$
$\underset{x\to \infty }{lim}f\left(x\right)=\underset{x\to \infty }{lim}{\left(\frac{ax+1}{bx+2}\right)}^x=\underset{x\to \infty }{lim}{\left(\frac{a+1/x}{b+2/x}\right)}^x=\underset{x\to \infty }{lim}{\left(\frac{a}{b}\right)}^x=\infty$
Case 3. $a=b$ form of the limit is $1^{\infty }$

$\underset{x\to \infty }{lim}f(x{)}^{g\left(x\right)}=e{}^{\underset{x\to \infty }{lim}\left[\right(f\left(x\right)-1\left)g\left(x\right)\right]}$

$\underset{x\to \infty }{lim}f\left(x\right)=\underset{x\to \infty }{lim}{\left(\frac{ax+1}{ax+2}\right)}^x=e^{\underset{x\to \infty }{lim}\left(\frac{ax+1}{ax+2}-1\right)x}=e^{\underset{x\to \infty }{lim}\left(\frac{-x}{ax+2}\right)}=e^{(-\frac{1}{a})}$