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Calculus Level 1

lim ⁡ n → 0 1 + n n = ? \large{\lim_{n\to0}{\sqrt[n]{1+n}}=?}

2 e 2 − e 2e^2-e 2 ln ⁡ 2 2\ln2 ln ⁡ 2 \ln2 e e e 2 e^2

Md Omur Faruque by MD Omur Faruque , 25, Bangladesh

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1 solution

Md Omur Faruque
Oct 10, 2015

lim ⁡ n → 0 1 + n n = lim ⁡ n → 0 ( 1 + n ) 1 n \lim_{n\to0}{\sqrt[n]{1+n}}=\lim_{n\to0}{(1+n)^{\dfrac1n}} = e ln ⁡ [ lim ⁡ n → 0 ( 1 + n ) 1 n ] =e^{\displaystyle \ln{[\lim_{n\to0}{(1+n)^{\frac1n}}]}} = e lim ⁡ n → 0 ln ⁡ ( 1 + n ) 1 n =e^{\displaystyle \lim_{n\to0}{\ln{(1+n)^{\frac1n}}}} = e lim ⁡ n → 0 ln ⁡ ( 1 + n ) n =e^{\displaystyle \lim_{n\to0}{\dfrac{\ln{(1+n)}}{n}}} = e lim ⁡ n → 0 1 1 + n 1 =e^{\displaystyle \lim_{n\to0}{\dfrac{\dfrac{1}{1+n}}{1}}} = e 1 = e =e^1=\color{#0C6AC7}{\boxed{e}}

However, it's only the definition of e e (slightly transformed though) from compound interest rate, which we don't need to prove. lim ⁡ n → 0 1 + n n = lim ⁡ n → 0 ( 1 + n ) 1 n \lim_{n\to0}{\sqrt[n]{1+n}}=\lim_{n\to0}{(1+n)^{\frac1n}} = lim ⁡ n → ∞ ( 1 + 1 n ) n = e =\lim_{n\to \infty}{\left(1+\frac1n\right)^n}=\color{#0C6AC7}{\boxed{e}}

For explanation please visit the Wikipedia page of e & Compound Interest .

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I accidentally published incomplete solution and I deleted it, so I have to post here, sorry! lim ⁡ n → 0 1 + n n = lim ⁡ n → 0 ( 1 + n ) 1 n \lim_{n\to0}\sqrt[n]{1+n} =\lim_{n\to0}(1+n)^{\frac1n} Let t = 1 n t = \dfrac1n , so since t → ∞ t\to\infty as n → 0 + n\to0^+ and t → − ∞ t\to-\infty as n → 0 − n\to0^- , we need to split limit into lim ⁡ n → 0 + 1 + n n \displaystyle \lim_{n\to0^+}\sqrt[n]{1+n} and lim ⁡ n → 0 − 1 + n n \displaystyle \lim_{n\to0^-}\sqrt[n]{1+n}

Using the fact that lim ⁡ n → ∞ ( 1 + x n ) n = e x \displaystyle \lim_{n\to\infty}\left(1+\dfrac x n\right)^n = e^x , we have: lim ⁡ n → 0 + 1 + n n = lim ⁡ t → ∞ ( 1 + 1 t ) t = e \lim_{n\to0^+}\sqrt[n]{1+n} = \lim_{t\to\infty} \left(1+\dfrac1t\right)^t = e

And lim ⁡ n → 0 − 1 + n n = lim ⁡ t → − ∞ ( 1 + 1 t ) t = lim ⁡ t → ∞ ( 1 + 1 − t ) − t = lim ⁡ t → ∞ 1 ( 1 + − 1 t ) t = 1 e − 1 = e \displaystyle \lim_{n\to0^-}\sqrt[n]{1+n} = \lim_{t\to-\infty} \left(1+\dfrac1t\right)^t = \lim_{t\to\infty} \left(1+\dfrac1{-t}\right)^{-t} = \lim_{t\to\infty}\dfrac1{ \left(1+\dfrac{-1}t\right)^t} = \dfrac1{e^{-1}} = e

Since both limits equal to e e , we have original limit equals to e \boxed{e} .

Micah Wood - 5 years, 6 months ago

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