# Evaluate $\displaystyle \large \lim_{x\,\to\,\infty}{\normalsize \bigg(1+\dfrac{4}{x}\bigg)^{\displaystyle 3x}}$ by Logarithms

It is evaluated that the limit of 1 plus 4 by x whole raised to the power of 3x as the value of $x$ approaches infinity is indeterminate as per the direct substitution method.

$\displaystyle \large \lim_{x\,\to\,\infty}{\normalsize \bigg(1+\dfrac{4}{x}\bigg)^{\displaystyle 3x}}$ $\,=\,$ $1^\infty$

The limit of the given function should not be indeterminate. So, its limit should be evaluated in another method. Let us learn how to find the limit of the function as $x$ tends to infinity.

### Release the function from Exponential notation

The formation of the given function in exponential notation is complex and it is the main reason for the indeterminate form. So, its complexity can be reduced only when the function is released from the exponential form.

Assume that $y$ $\,=\,$ $\bigg(1+\dfrac{4}{x}\bigg)^{\displaystyle 3x}$

Now, take natural logarithm on both sides of the equation in order to release the given function from exponential notation.

$\implies$ $\log_{e}{(y)}$ $\,=\,$ $\log_{e}{\bigg(1+\dfrac{4}{x}\bigg)^{\displaystyle 3x}}$

On right hand side of the equation, the logarithm of $1$ plus $4$ by $x$ whole power $3x$ can be simplified by the power rule of logarithms.

$\,\,\,\therefore\,\,\,\,\,\,$ $\log_{e}{(y)}$ $\,=\,$ $3x \times \log_{e}{\bigg(1+\dfrac{4}{x}\bigg)}$

### Minimize the complexity of Logarithmic equation

The given function in exponential notation is successfully written as a logarithmic equation. It is time to focus on simplifying the logarithmic equation.

$\log_{e}{(y)}$ $\,=\,$ $3x \times \log_{e}{\bigg(1+\dfrac{4}{x}\bigg)}$

On left hand side of the log equation, there is nothing to simplify but we can do on right hand side of the equation. The literal factor $3x$ cannot be multiplied with logarithmic function but the logarithmic function can be expanded by the expansion of logarithm of a binomial in one variable.

$\log_{e}{(1+z)}$ $\,=\,$ $z$ $-$ $\dfrac{z^2}{2}$ $+$ $\dfrac{z^3}{3}$ $-$ $\dfrac{z^4}{4}$ $+$ $\cdots$

According to the logarithmic expansion, the logarithm of $1$ plus $4$ by $x$ can be expanded by substituting the $z$ by $4$ by $x$.

$\implies$ $\log_{e}{(y)}$ $\,=\,$ $3x \times \bigg(\dfrac{4}{x}$ $-$ $\dfrac{\Big(\dfrac{4}{x}\Big)^2}{2}$ $+$ $\dfrac{\Big(\dfrac{4}{x}\Big)^3}{3}$ $-$ $\dfrac{\Big(\dfrac{4}{x}\Big)^4}{4}$ $+$ $\cdots\bigg)$

The quotient of $4$ by $x$ with different exponents can be simplified by the power of a quotient rule.

$\implies$ $\log_{e}{(y)}$ $\,=\,$ $3x \times \bigg(\dfrac{4}{x}$ $-$ $\dfrac{\dfrac{4^2}{x^2}}{2}$ $+$ $\dfrac{\dfrac{4^3}{x^3}}{3}$ $-$ $\dfrac{\dfrac{4^4}{x^4}}{4}$ $+$ $\cdots\bigg)$

Now, simplify the expression on the right hand side of the equation.

$\implies$ $\log_{e}{(y)}$ $\,=\,$ $3x \times \bigg(\dfrac{4}{x}$ $-$ $\dfrac{\dfrac{16}{x^2}}{2}$ $+$ $\dfrac{\dfrac{64}{x^3}}{3}$ $-$ $\dfrac{\dfrac{256}{x^4}}{4}$ $+$ $\cdots\bigg)$

$\implies$ $\log_{e}{(y)}$ $\,=\,$ $3x \times \bigg(\dfrac{4}{x}$ $-$ $\dfrac{16}{x^2 \times 2}$ $+$ $\dfrac{64}{x^3 \times 3}$ $-$ $\dfrac{256}{x^4 \times 4}$ $+$ $\cdots\bigg)$

$\implies$ $\log_{e}{(y)}$ $\,=\,$ $3x \times \bigg(\dfrac{4}{x}$ $-$ $\dfrac{\cancel{16}}{x^2 \times \cancel{2}}$ $+$ $\dfrac{64}{x^3 \times 3}$ $-$ $\dfrac{\cancel{256}}{x^4 \times \cancel{4}}$ $+$ $\cdots\bigg)$

$\implies$ $\log_{e}{(y)}$ $\,=\,$ $3x \times \bigg(\dfrac{4}{x}$ $-$ $\dfrac{8}{x^2}$ $+$ $\dfrac{64}{3x^3}$ $-$ $\dfrac{64}{x^4}$ $+$ $\cdots\bigg)$

The literal factor $3x$ can be distributed over the sum or difference of terms by the distributive property.

$\implies$ $\log_{e}{(y)}$ $\,=\,$ $3x \times \dfrac{4}{x}$ $-$ $3x \times \dfrac{8}{x^2}$ $+$ $3x \times \dfrac{64}{3x^3}$ $-$ $3x \times \dfrac{64}{x^4}$ $+$ $\cdots$

Now, simplify each term on the right hand side of the equation to complete the simplification process of the equation.

$\implies$ $\log_{e}{(y)}$ $\,=\,$ $\dfrac{3x \times 4}{x}$ $-$ $\dfrac{3x \times 8}{x^2}$ $+$ $\dfrac{3x \times 64}{3x^3}$ $-$ $\dfrac{3x \times 64}{x^4}$ $+$ $\cdots$

$\implies$ $\log_{e}{(y)}$ $\,=\,$ $\dfrac{3 \times x \times 4}{x}$ $-$ $\dfrac{3 \times x \times 8}{x^2}$ $+$ $\dfrac{3 \times x \times 64}{3 \times x^3}$ $-$ $\dfrac{3 \times x \times 64}{x^4}$ $+$ $\cdots$

$\implies$ $\log_{e}{(y)}$ $\,=\,$ $\dfrac{3 \times \cancel{x} \times 4}{\cancel{x}}$ $-$ $\dfrac{3 \times \cancel{x} \times 8}{\cancel{x^2}}$ $+$ $\dfrac{\cancel{3} \times \cancel{x} \times 64}{\cancel{3} \times \cancel{x^3}}$ $-$ $\dfrac{3 \times \cancel{x} \times 64}{\cancel{x^4}}$ $+$ $\cdots$

$\implies$ $\log_{e}{(y)}$ $\,=\,$ $3 \times 4$ $-$ $\dfrac{3 \times 8}{x}$ $+$ $\dfrac{64}{x^2}$ $-$ $\dfrac{3 \times 64}{x^3}$ $+$ $\cdots$

$\,\,\,\therefore\,\,\,\,\,\,$ $\log_{e}{(y)}$ $\,=\,$ $12$ $-$ $\dfrac{24}{x}$ $+$ $\dfrac{64}{x^2}$ $-$ $\dfrac{192}{x^3}$ $+$ $\cdots$

### Evaluate the Limit of the exponential function

The simplification process of the logarithmic equation is completed, and it is time to calculate the limit on both sides of the equation as the value of $x$ approaches infinity.

$\implies$ $\displaystyle \large \lim_{x\,\to\,\infty}{\normalsize \log_{e}{(y)}}$ $\,=\,$ $\displaystyle \large \lim_{x\,\to\,\infty}{\normalsize \bigg(12}$ $-$ $\dfrac{24}{x}$ $+$ $\dfrac{64}{x^2}$ $-$ $\dfrac{192}{x^3}$ $+$ $\cdots\bigg)$

On the right hand side of the equation, use the direct substitution method to find the limit of the infinite series as the value of $x$ is closer to infinity.

$\implies$ $\displaystyle \large \lim_{x\,\to\,\infty}{\normalsize \log_{e}{(y)}}$ $\,=\,$ $12$ $-$ $\dfrac{24}{\infty}$ $+$ $\dfrac{64}{\infty^2}$ $-$ $\dfrac{192}{\infty^3}$ $+$ $\cdots$

Now, simplify the infinite series on the right hand side of the equation to find its value.

$\implies$ $\displaystyle \large \lim_{x\,\to\,\infty}{\normalsize \log_{e}{(y)}}$ $\,=\,$ $12$ $-$ $\dfrac{24}{\infty}$ $+$ $\dfrac{64}{\infty}$ $-$ $\dfrac{192}{\infty}$ $+$ $\cdots$

$\implies$ $\displaystyle \large \lim_{x\,\to\,\infty}{\normalsize \log_{e}{(y)}}$ $\,=\,$ $12$ $-$ $0$ $+$ $0$ $-$ $0$ $+$ $\cdots$

$\,\,\,\therefore\,\,\,\,\,\,$ $\displaystyle \large \lim_{x\,\to\,\infty}{\normalsize \log_{e}{(y)}}$ $\,=\,$ $12$

It is time to find the limit of the expression on the left-hand side of the equation. The logarithmic function is defined in terms of $y$. So, it should be replaced by its actual value in terms of $x$.

$\implies$ $\displaystyle \large \lim_{x\,\to\,\infty}{\normalsize \log_{e}{\Bigg(\bigg(1+\dfrac{4}{x}\bigg)^{\displaystyle 3x}\Bigg)}}$ $\,=\,$ $12$

The function on the left-hand side of the equation is a composite function. So, its limit can be evaluated by the composite limit rule.

$\implies$ $\log_{e}{\Bigg(\displaystyle \large \lim_{x\,\to\,\infty}{\normalsize \bigg(1+\dfrac{4}{x}\bigg)^{\displaystyle 3x}\Bigg)}}$ $\,=\,$ $12$

It is time to eliminate the involvement of logarithm and it can be done by the relationship between the logarithmic and exponential functions.

$\,\,\,\therefore\,\,\,\,\,\,$ $\displaystyle \large \lim_{x\,\to\,\infty}{\normalsize \bigg(1+\dfrac{4}{x}\bigg)^{\displaystyle 3x}}$ $\,=\,$ $e^{12}$

A best free mathematics education website for students, teachers and researchers.

###### Maths Topics

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

###### Maths Problems

Learn how to solve the maths problems in different methods with understandable steps.

Learn solutions

###### Subscribe us

You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites.

Copyright © 2012 - 2022 Math Doubts, All Rights Reserved