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

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$e^{2}$$\,\,\left(\approx 7.3890560989306495\right) Got another answer? Verify it here ## 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)$
SnapXam A2

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ln
log
log
lim
d/dx
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|◻|
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>=
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sin
cos
tan
cot
sec
csc

asin
acos
atan
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acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
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acoth
asech
acsch

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