Question

Evaluate lim x tends to infinity (ax+1/ax-1)^x

Answer 1

$\lim_{x \to \infty}$ $(\frac{ax+1}{ax-1})^{x}$= $e^\frac{2}{a}$

Step-by-step explanation:

Given: a function $(\frac{ax+1}{ax-1})^{x}$

To find: the value of the function as $_{x \to \infty}$

Check the form of the limit:

Dividing both numerator and denominator terms inside the parenthesis by $x$ we get:

$(\frac{a+\frac{1}{x}}{a- \frac{1}{x} }) ^{x}$

As $_{x \to \infty}$,   $_{\frac{1}{x} \to 0}$

⇒  $\lim_{x \to \infty} (\frac{a+\frac{1}{x}}{a- \frac{1}{x} }) ^{x}$$=1^{\infty}$

The limit is of the form $1^{\infty}$ which is indeterminate.

How to find the limit?

Let $f(x)$ and $g(x)$ be two functions such that;

$\lim_{x \to \infty} f(x)=1$   and

$\lim_{x \to \infty} g(x)=\infty$

Then we have:

$\lim_{x \to \infty} f(x)^{g(x)}$ $=e^{ \lim_{x \to \infty} (f(x)-1)*g(x)}$

Using this formula for the given question we get:

$\lim_{x\to \infty}$  $(\frac{ax+1}{ax-1})^{x}$$= e^{ \lim_{x \to \infty}(\frac{ax+1}{ax-1} -1)*x }$

                        $=e^{ \lim_{x \to \infty} (\frac{ax+1-ax+1}{ax-1})*x}$

                        $=e^{ \lim_{x \to \infty} \frac{2x}{ax-1} }$

                        $=e^{ \lim_{x \to \infty} \frac{2}{a-\frac{1}{x} } }$

                       $=e^{\frac{2}{a} }$                                 $[ \lim_{x \to \infty} \frac{1}{x}=0 ]$

Hence, the required limit is $e^\frac{2}{a}$