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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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Final Answer

$e^{2}$
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Step-by-step Solution

Problem to solve:

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

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 Try to guess Step 3. Or become premium for the price of a latte.
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

$\frac{\ln\left(1+0\right)}{\frac{1}{\infty }}$

Add the values $1$ and $0$

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

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 Try to guess Step 6. Or become premium for the price of a latte.

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 ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

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

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

$\frac{1}{1+\frac{2}{x}}\frac{0x-2\frac{d}{dx}\left(x\right)}{x^2}$

Any expression multiplied by $0$ is equal to $0$

$\frac{1}{1+\frac{2}{x}}\frac{0-2\frac{d}{dx}\left(x\right)}{x^2}$

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

$\frac{1}{1+\frac{2}{x}}\frac{0-2}{x^2}$

$x+0=x$, where $x$ is any expression

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

Multiplying fractions $\frac{1}{1+\frac{2}{x}} \times \frac{-2}{x^2}$

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

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

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

Multiplying the fraction by $x^2$

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

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

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

Simplify the fraction by $x$

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

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 ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

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

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

$\frac{0x-\frac{d}{dx}\left(x\right)}{x^2}$

Any expression multiplied by $0$ is equal to $0$

$\frac{0-\frac{d}{dx}\left(x\right)}{x^2}$

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

$\frac{0-1}{x^2}$

$x+0=x$, where $x$ is any expression

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

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

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

Simplify the fraction by $x$

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

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

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

$e^{\lim_{x\to\infty }\left(\frac{2x}{2+x}\right)}$
9 Try to guess Step 9. Or become premium for the price of a latte.
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 Try to guess Step 12. Or become premium for the price of a latte.
13

Calculate the power $e^{2}$

$e^{2}$

Final Answer

$e^{2}$

Numeric Answer

$7.389056$

Explore different ways to solve this problem

Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more

Limits by direct substitutionLimits by L'Hôpital's ruleLimits by factoringLimits by rationalizing

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.
(◻)
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e
π
ln
log
log
lim
d/dx
Dx
|◻|
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

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