Question

Consider the function \[f\left( x \right)={{\left( \dfrac{ax+1}{bx+2} \right)}^{x}}\] where \[a,b> 0\] then \[\displaystyle \lim_{x \to \infty }f\left( x \right)\]\[\_\_\_\_\_\_\]
This question has multiple correct options.
A. exists for all values of \[a\] and \[b\]
B. is zero for \[a< b\]
C. is non-existent for \[a> b\]
D. is \[{{e}^{\left( \dfrac{-1}{a} \right)}}\] or \[{{e}^{\left( \dfrac{-1}{b} \right)}}\] if \[a=b\]

Answer 1

Hint: In order to solve the given problem, we will be checking out for each and every option given by applying the limits to the given function \[f\left( x \right)={{\left( \dfrac{ax+1}{bx+2} \right)}^{x}}\] and computing it. After obtaining the answer, we will be checking if the given statement holds for the given option.

Complete step-by-step solution:
Now let us have a brief about functions and limits. Generally, the limit of a function is nothing but it is the value that approaches as the input approaches some value. When the function tends to infinity, the function does not have any kind of limit as the function does not get closer to any real number as the value of \[x\] keeps getting larger.
Now let us solve our given function \[f\left( x \right)={{\left( \dfrac{ax+1}{bx+2} \right)}^{x}}\].
Considering the first option, i.e. A. exists for all values of \[a\] and \[b\]
As the limit tends to infinity, we can say that it does not exist for all the values of \[a\] and \[b\]
Hence, option A isn't the correct option.
Considering the second option, i.e. is zero for \[a< b\]
When \[a< b\],
\[\begin{align}
  & \Rightarrow \displaystyle \lim_{x \to \infty }f\left( x \right)=\displaystyle \lim_{x \to \infty }{{\left( \dfrac{ax+1}{bx+2} \right)}^{x}} \\
 & \Rightarrow \displaystyle \lim_{x \to \infty }{{\left( \dfrac{a+\dfrac{1}{x}}{b+\dfrac{1}{x}} \right)}^{x}}=\displaystyle \lim_{x \to \infty }{{\left( \dfrac{a}{b} \right)}^{x}}=0 \\
\end{align}\]
\[\therefore \] Option B is correct.
Considering the third option, i.e. is non-existent for \[a> b\]
When \[a> b\],
\[\begin{align}
  & \Rightarrow \displaystyle \lim_{x \to \infty }f\left( x \right)=\displaystyle \lim_{x \to \infty }{{\left( \dfrac{ax+1}{bx+2} \right)}^{x}} \\
 & \Rightarrow \displaystyle \lim_{x \to \infty }{{\left( \dfrac{ax+\dfrac{1}{x}}{bx+\dfrac{1}{x}} \right)}^{x}}=\displaystyle \lim_{x \to \infty }{{\left( \dfrac{a}{b} \right)}^{x}}=\infty \\
\end{align}\]
Hence option C is correct.
Now, let us consider the fourth option, i.e. is \[{{e}^{\left( \dfrac{-1}{a} \right)}}\] or \[{{e}^{\left( \dfrac{-1}{b} \right)}}\] if \[a=b\]
When \[a=b\], it is of the form \[{{1}^{\infty }}\]
The general form to compute this limit is, \[\displaystyle \lim_{x \to \infty }f{{\left( x \right)}^{g\left( x \right)}}={{e}^{\displaystyle \lim_{x \to \infty }\left[ \left( f\left( x \right)-1 \right)g\left( x \right) \right]}}\]
Now let us compute our function-
$ \Rightarrow \displaystyle \lim_{x \to \infty }f\left( x \right)=\displaystyle \lim_{x \to \infty }{{\left( \dfrac{ax+1}{bx+2} \right)}^{x}} $
$\Rightarrow {{e}^{\displaystyle \lim_{x \to \infty }\left( \dfrac{ax+1}{bx+2} \right)x}}={{e}^{\displaystyle \lim_{x \to \infty }\left( \dfrac{-x}{ax+2} \right)={{e}^{\left( \dfrac{-1}{a}\right)}}}} $
Hence, option D also satisfies.
\[\therefore \] The options that satisfy the given function with given limit conditions are B,C,D.

Note: We must always have a note that if the limit is zero, then the function would be undefined. We also observe that we will be obtaining a variety of answers. In order to know the correct one, we must use the L-Hospital’s rule to compute them.