Question

What is the limit of ${{\left( 1+2x \right)}^{\dfrac{1}{x}}}$ as ‘x’ approaches infinity ?

Answer 1

To find the limiting value of any given expression, we will first of all determine what type of indeterminate form the expression represents and then proceed accordingly. In the problem given to us, the indeterminate form is infinity to the power of zero. This can be solved by converting the expression into an exponential function raised to some power.

Complete step-by-step solution:
Let us represent our given expression with the term ‘y’, such that our problem now becomes:
$\Rightarrow y=\displaystyle \lim_{x \to \infty }{{\left[ 1+2x \right]}^{\dfrac{1}{x}}}$
First of all we will change the indeterminate form of the given expression. This can be done by changing the function into an exponential function. This is done as follows:
\begin{aligned} & \Rightarrow y=\displaystyle \lim_{x \to \infty }\left[ {{\left\\{ {{e}^{\ln \left( 1+2x \right)}} \right\\}}^{\dfrac{1}{x}}} \right] \\\ & \Rightarrow y=\displaystyle \lim_{x \to \infty }\left[ {{e}^{\dfrac{\ln \left( 1+2x \right)}{x}}} \right] \\\ & \Rightarrow y=\left[ {{e}^{\displaystyle \lim_{x \to \infty }\dfrac{\ln \left( 1+2x \right)}{x}}} \right] \\\ \end{aligned}
Now, we can see that the power of the exponent is an indeterminate form of $\dfrac{\infty }{\infty }$. This indeterminate form can be easily solved using the L’Hospital’s rule. This rule states that whenever a function has $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ indeterminate form, we can calculate the limiting value of expression by successively differentiating the numerator and denominator until the indeterminate form is lost.
This can be done to the power $\left[ say,\text{ g}\left( x \right) \right]$ of our exponent as follows:
$\begin{aligned} & \Rightarrow g\left( x \right)=\displaystyle \lim_{x \to \infty }\dfrac{d\left( \dfrac{\ln \left( 1+2x \right)}{dx} \right)}{\dfrac{dx}{dx}} \\\ & \Rightarrow g\left( x \right)=\displaystyle \lim_{x \to \infty }\dfrac{\dfrac{2}{1+2x}}{1} \\\ & \Rightarrow g\left( x \right)=\displaystyle \lim_{x \to \infty }\dfrac{2}{1+2x} \\\ & \therefore g\left( x \right)=0 \\\ \end{aligned}$
Now, using this limiting value of $g\left( x \right)$ in our original expression, we get:
$\begin{aligned} & \Rightarrow y={{e}^{0}} \\\ & \therefore y=1 \\\ \end{aligned}$
Thus, the final limiting value of our expression comes out to be 1.
Hence, the limiting value of ${{\left( 1+2x \right)}^{\dfrac{1}{x}}}$ as ‘x’ approaches infinity comes out to be 1.

Note: There are some instances when L’Hospital’s rule does not give us any result or one can say leads us to an infinite number of differentials. In such an event, we should try solving the problem using basic factorization and taking out common factors. And if, this does not work, we should try changing the indeterminate form of the expression.