Question

Lim \[\left( {x \to \inf } \right)\] of \[\left[ {{{\left( {1 + \dfrac{1}{x}} \right)}^x}} \right] \] how to explain it via calculator explanation and/or implications/applications?

Answer 1

Hint: To solve the given expression consists of limits with \[\left( {x \to \inf } \right)\] i.e., \[x \to \infty \] , in which we need to apply limits such that we must use L'Hopital's rule and take derivatives of both the numerator and the denominator. Hence, using a calculator, consider x values to find the value of the given expression as we know that, when \[x \to \infty \] the limit is undefined.

Complete step-by-step answer:
Given,
 \[y = \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \left( {\dfrac{1}{x}} \right)} \right)^x}\]
Apply ln on both sides of the expression:
 \[\ln y = \mathop {\lim }\limits_{x \to \infty } \ln {\left( {1 + \left( {\dfrac{1}{x}} \right)} \right)^x}\]
Now, the power term x is written as:
 \[\ln y = \mathop {\lim }\limits_{x \to \infty } x\ln \left( {1 + \left( {\dfrac{1}{x}} \right)} \right)\]
 \[ \Rightarrow \ln y = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\ln \left( {1 + \left( {\dfrac{1}{x}} \right)} \right)}}{{{x^{ - 1}}}}\] ……………… 1
When \[x \to \infty \] , the limit is undefined. As such, we need to use L'Hopital's rule and take derivatives of both the numerator and the denominator:
L'Hopital's rule says that if \[\mathop {\lim }\limits_{x \to a} f\left( x \right) = 0 = \mathop {\lim }\limits_{x \to a} g\left( x \right)\]
Then,
 \[\mathop {\lim }\limits_{x \to a} \left( {\dfrac{{f\left( x \right)}}{{g\left( x \right)}}} \right) = \mathop {\lim }\limits_{x \to a} \left( {\dfrac{{f'\left( x \right)}}{{g'\left( x \right)}}} \right)\]
Hence, applying this we get equation 1 as:
 \[\ln y = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\ln \left( {1 + \left( {\dfrac{1}{x}} \right)} \right)}}{{{x^{ - 1}}}}\] ………………….. 2
Now, we need to find the derivative of the equation 2 as:
We know that, the derivative of ln x is: \[\dfrac{d}{{dx}}\left( {\ln x} \right) = \dfrac{1}{{x'}}\] , hence apply to the numerator terms and also \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n.{x^{n - 1}}\] , hence applying to the denominator terms we get:
 \[ \Rightarrow \ln y = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\left( {\dfrac{1}{{1 + \left( {\dfrac{1}{x}} \right)}}} \right)\left( {0 - 1{x^{ - 2}}} \right)}}{{ - 1{x^{ - 2}}}}\]
Simplifying the terms as there is a common term involved of \[ - 1{x^{ - 2}}\] in both numerator and denominator, hence we get:
 \[ \Rightarrow \ln y = \mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{1}{{1 + \left( {\dfrac{1}{x}} \right)}}} \right)\]
Now, we need to plug in \[\infty \] for x i.e., applying the limits:
 \[\ln y = \left( {\dfrac{1}{{1 + 0}}} \right)\]
Hence, simplifying we get:
 \[\ln y = \left( {\dfrac{1}{1}} \right)\]
 \[ \Rightarrow \ln y = 1\] …………………. 2
Now, we need to use the exponential rule for ‘e’ i.e., \[{e^{\ln x}} = x\] , hence here from equation 2 we have:
 \[{e^{\ln y}} = {e^1}\]
 \[ \Rightarrow y = e\]
Applying it to the given expression as:
 \[y = e = \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \left( {\dfrac{1}{x}} \right)} \right)^x}\]
 \[ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {\left( {1 + \left( {\dfrac{1}{x}} \right)} \right)^x} = e\]
If you use your calculator, you need to begin with \[x = 1\] and then calculate the value of the function. Then you can increase the value of x as shown in the table below.

\[x\]	\[g\left( x \right) = {\left( {1 + \dfrac{1}{x}} \right)^x}\]
1	\[{\left( {1 + \dfrac{1}{1}} \right)^1} = 2\]
10	\[{\left( {1 + \dfrac{1}{{10}}} \right)^{10}} = 2.593742\]
1000	\[{\left( {1 + \dfrac{1}{{1000}}} \right)^{1000}} = 2.716924\]
\[1 \times {10^6}\]	\[{\left( {1 + \dfrac{1}{{1 \times {{10}^6}}}} \right)^{1 \times {{10}^6}}} = 2.718280\]
\[1 \times {10^{10}}\]	\[{\left( {1 + \dfrac{1}{{1 \times {{10}^{10}}}}} \right)^{1 \times {{10}^{10}}}} = 2.718282\]

Hence, you can see the trend that the value of the function, as x increases, gets closer and closer to the value of e i.e., \[e \simeq 2.718\] .
The applications of this are in calculating values that other types of series converge to in calculus. In real life, as an Electronics Engineer, we know that it is used in estimating what is called the "Steady-State Error" of functions in Control Systems Engineering problems, Reliability Engineering Problems, Current and Voltage dissipation of electronic components, etc.
There are probably applications in other fields such as financial calculations for rates of return of certain investments, and analysis of performance over time with steadily reducing rates of return; as well as in other scientific fields.

Note: ln stands for natural logarithm with a base of e, in which e is a natural base and is approximately equal to 2.718. Hence, \[y = {a^x}\] is in exponential form and \[x = {\log _a}y\] is in logarithmic form. We must note that \[\ln \left( 1 \right) = 0\] , this implies that \[{e^0} = 1\] which makes sense. You can also see this equivalence by making both sides exponents of \[e\] .