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Find the limit of $\left(1+\frac{2}{x}\right)^x$ as $x$ approaches $\infty $

Step-by-step Solution

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Solution

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Final Answer

$e^{2}$$\,\,\left(\approx 7.3890560989306495\right)$
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Step-by-step Solution

Problem to solve:

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

Specify the 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: $\displaystyle{\lim_{x\to a}f(x)^{g(x)} = \lim_{x\to a}f(x)^{\displaystyle\lim_{x\to a}g(x)}}$

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

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

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

$e^{\lim_{x\to\infty }\left(\frac{-2x}{-\left(2+x\right)}\right)}$
8

Cancel the negative coefficients in $\frac{-2x}{-\left(2+x\right)}$

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

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

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

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

$\frac{\infty }{\infty }$
10

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 or more 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 $1$

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

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

$e^{\lim_{x\to\infty }\left(2\right)}$
12

The limit of a constant is just the constant

$e^{2}$

Final Answer

$e^{2}$$\,\,\left(\approx 7.3890560989306495\right)$
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(◻)
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/
÷
◻2
◻◻
√◻
√

◻√◻
◻√
∞
e
π
ln
log
log◻
lim
d/dx
D□x
∫
∫◻
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

Tips on how to improve your answer:

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

Main topic:

Limits to Infinity

Related Formulas:

7. See formulas

Time to solve it:

~ 0.12 s

Related topics:

Limits to InfinityLimitsBasic Differentiation RulesSum Rule of DifferentiationConstant Rule for DifferentiationQuotient Rule of differentiationLimits by L'Hôpital's rule

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