Limited yet powerful (III)

Calculus Level 3

lim ⁡ x → 0 + ∣ x + 1 x − 1 ∣ 1 / x \large \lim_{x\to0^{+}} \left|\dfrac{x+1}{x-1} \right|^{1/x}

If the value of the limit above is equal to e N e^N , find the value of N N .


The answer is 2.

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Noel Lo by Noel Lo , 24, Singapore

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1 solution

Noel Lo
Nov 30, 2015

lim ⁡ x → 0 ∣ x + 1 x − 1 ∣ 1 x = e x p ( lim ⁡ x → 0 l n ∣ x + 1 x − 1 ∣ 1 x ) = e x p ( lim ⁡ x → 0 1 x l n ∣ x + 1 x − 1 ∣ ) = e x p ( lim ⁡ x → 0 1 x ( l n ∣ x + 1 ∣ − l n ∣ x − 1 ∣ ) ) \lim_{x\rightarrow0}|\frac{x+1}{x-1}|^{\frac{1}{x}} =exp(\lim_{x\rightarrow0}ln|\frac{x+1}{x-1}|^{\frac{1}{x}} )=exp(\lim_{x\rightarrow0}{\frac{1}{x}} ln|\frac{x+1}{x-1}|) =exp(\lim_{x\rightarrow0}{\frac{1}{x}} (ln|x+1|-ln|x-1|))

= e x p ( lim ⁡ x → 0 1 x + 1 − 1 x − 1 1 ) = e x p lim ⁡ x → 0 − 2 ( x + 1 ) ( x − 1 ) = e x p − 2 1 ( − 1 ) = e x p ( 2 ) = e 2 =exp(\lim_{x\rightarrow0} \frac{ \frac{1}{x+1}- \frac{1}{x-1}}{1})=exp\lim_{x\rightarrow0} \frac{-2}{(x+1)(x-1)}=exp\frac{-2}{1(-1)}=exp(2)=e^{2}

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Is this using L'Hopital's?

Kevin Erdiza - 5 years, 6 months ago

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Yes. The second line contains the derivative of the last step on the first line.

Krutarth Patel - 3 months, 1 week ago

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