Step-by-step Solution

Find the limit $\lim_{x\to\infty }\left(\left(1+\frac{2}{x}\right)^x\right)$

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Final Answer

$e^{2}$$\,\,\left(\approx 7.3890560989306495\right)$

Step-by-step Solution

Problem to solve:

$\lim_{x\to\infty}\left(1+\frac{2}{x}\right)^x$

Solving method

1

Rewrite the limit using the identity: $a^x=e^{x\ln\left(a\right)}$

$\lim_{x\to\infty }\left(e^{x\ln\left(1+\frac{2}{x}\right)}\right)$
2

Apply the power rule of limits: $\displaystyle{\lim_{x\to a}f(x)^{g(x)} = \lim_{x\to a}f(x)^{\displaystyle\lim_{x\to a}g(x)}}$

$\lim_{x\to\infty }\left(e\right)^{\lim_{x\to\infty }\left(x\ln\left(1+\frac{2}{x}\right)\right)}$
3

The limit of a constant is just the constant

$e^{\lim_{x\to\infty }\left(x\ln\left(1+\frac{2}{x}\right)\right)}$
4

Rewrite the product inside the limit as a fraction

$\lim_{x\to \infty }\left(\frac{\ln\left(1+\frac{2}{x}\right)}{\frac{1}{x}}\right)$

Plug in the value $\infty $ into the limit

$\frac{\ln\left(1+\frac{2}{\infty }\right)}{\frac{1}{\infty }}$

Any expression divided by infinity is equal to zero

$\frac{\ln\left(1\right)}{\frac{1}{\infty }}$

Calculating the natural logarithm of $1$

$\frac{0}{\frac{1}{\infty }}$

Any expression divided by infinity is equal to zero

$\frac{0}{0}$
5

If we directly evaluate the limit $\lim_{x\to \infty }\left(\frac{\ln\left(1+\frac{2}{x}\right)}{\frac{1}{x}}\right)$ as $x$ tends to $\infty $, we can see that it gives us an indeterminate form

$\frac{0}{0}$
6

We can solve this limit by applying L'Hôpital's rule, which consists of calculating the derivative of both the numerator and the denominator separately

$\lim_{x\to \infty }\left(\frac{\frac{d}{dx}\left(\ln\left(1+\frac{2}{x}\right)\right)}{\frac{d}{dx}\left(\frac{1}{x}\right)}\right)$

Find the derivative of the numerator

$\frac{d}{dx}\left(\ln\left(1+\frac{2}{x}\right)\right)$

The derivative of the natural logarithm of a function is equal to the derivative of the function divided by that function. If $f(x)=ln\:a$ (where $a$ is a function of $x$), then $\displaystyle f'(x)=\frac{a'}{a}$

$\frac{1}{1+\frac{2}{x}}\frac{d}{dx}\left(1+\frac{2}{x}\right)$

The derivative of a sum of two functions is the sum of the derivatives of each function

$\frac{1}{1+\frac{2}{x}}\left(\frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(\frac{2}{x}\right)\right)$

The derivative of the constant function ($1$) is equal to zero

$\frac{1}{1+\frac{2}{x}}\frac{d}{dx}\left(\frac{2}{x}\right)$

Apply the quotient rule for differentiation, which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

$\frac{1}{1+\frac{2}{x}}\frac{x\frac{d}{dx}\left(2\right)-2\frac{d}{dx}\left(x\right)}{x^2}$

The derivative of the constant function ($2$) is equal to zero

$\frac{1}{1+\frac{2}{x}}\frac{-2\frac{d}{dx}\left(x\right)}{x^2}$

The derivative of the linear function is equal to $1$

$\frac{1}{1+\frac{2}{x}}\frac{-2}{x^2}$

Combine $1+\frac{2}{x}$ in a single fraction

$\frac{x}{2+x}\frac{-2}{x^2}$

Multiplying fractions $\frac{x}{2+x} \times \frac{-2}{x^2}$

$\frac{-2x}{x^2\left(2+x\right)}$

Simplify the fraction by $x$

$\frac{-2}{x\left(2+x\right)}$

Find the derivative of the denominator

$\frac{d}{dx}\left(\frac{1}{x}\right)$

Apply the quotient rule for differentiation, which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

$\frac{x\frac{d}{dx}\left(1\right)-\frac{d}{dx}\left(x\right)}{x^2}$

The derivative of the constant function ($1$) is equal to zero

$\frac{-\frac{d}{dx}\left(x\right)}{x^2}$

The derivative of the linear function is equal to $1$

$\frac{-1}{x^2}$

Divide fractions $\frac{\frac{-2}{x\left(2+x\right)}}{\frac{-1}{x^2}}$ with Keep, Change, Flip: $\frac{a}{b}\div c=\frac{a}{b}\div\frac{c}{1}=\frac{a}{b}\times\frac{1}{c}=\frac{a}{b\cdot c}$

$e^{\lim_{x\to\infty }\left(\frac{2x^2}{x\left(2+x\right)}\right)}$

Simplify the fraction by $x$

$e^{\lim_{x\to\infty }\left(\frac{2x}{2+x}\right)}$
7

After deriving both the numerator and denominator, the limit results in

$e^{\lim_{x\to\infty }\left(\frac{2x}{2+x}\right)}$

Plug in the value $\infty $ into the limit

$\frac{2\infty }{2+\infty }$

Any expression multiplied by infinity tends to infinity

$\frac{\infty }{2+\infty }$

Infinity plus any algebraic expression is equal to infinity

$\frac{\infty }{\infty }$
8

If we directly evaluate the limit $\lim_{x\to \infty }\left(\frac{2x}{2+x}\right)$ as $x$ tends to $\infty $, we can see that it gives us an indeterminate form

$\frac{\infty }{\infty }$
9

We can solve this limit by applying L'Hôpital's rule, which consists of calculating the derivative of both the numerator and the denominator separately

$\lim_{x\to \infty }\left(\frac{\frac{d}{dx}\left(2x\right)}{\frac{d}{dx}\left(2+x\right)}\right)$

Find the derivative of the numerator

$\frac{d}{dx}\left(2x\right)$

The derivative of the linear function times a constant, is equal to the constant

$2$

Find the derivative of the denominator

$\frac{d}{dx}\left(2+x\right)$

The derivative of a sum of two functions is the sum of the derivatives of each function

$\frac{d}{dx}\left(2\right)+\frac{d}{dx}\left(x\right)$

The derivative of the constant function ($2$) is equal to zero

$\frac{d}{dx}\left(x\right)$

The derivative of the linear function is equal to $1$

$1$

Divide $2$ by $1$

$e^{\lim_{x\to\infty }\left(2\right)}$
10

After deriving both the numerator and denominator, the limit results in

$e^{\lim_{x\to\infty }\left(2\right)}$
11

The limit of a constant is just the constant

$e^{2}$

Final Answer

$e^{2}$$\,\,\left(\approx 7.3890560989306495\right)$