It's x 2 x^2 in the Denominator

Calculus Level 2

lim x 0 e x + e x 2 x 2 = ? \large \lim_{x \to 0} \dfrac{e^x + e^{-x} - 2}{x^2} = \, ?

0 1 2 \infty

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2 solutions

1.- lim x 0 e x + e x 2 x 2 = lim x 0 1 + x + x 2 2 ! + 1 x + ( x ) 2 2 ! 2 x 2 = 1 \lim_{x \to 0} \frac{e^x + e^{-x} - 2}{x^2} = \lim_{x \to 0} \frac{1 + x + \frac{x^2}{2!} + 1 - x + \frac{(-x)^2}{2!} - 2}{x^2} = 1 I have used e x e^x ~ 1 + x + x 2 2 ! 1 + x + \frac{x^2}{2!} when x 0 x \to 0 applying McLaurin series

2.- lim x 0 e x + e x 2 x 2 = lim x 0 2 cos i x 2 x 2 = lim x 0 2 ( cos i x 1 ) x 2 = lim x 0 2 ( ( i x ) 2 2 ! ) x 2 = 1 \lim_{x \to 0} \frac{e^x + e^{-x} - 2}{x^2} = \lim_{x \to 0} \frac{2 \cos ix - 2}{x^2} = \lim_{x \to 0} \frac{2(\cos ix -1)}{x^2} = \lim_{x \to 0} \frac{2(\frac{-(ix)^2}{2!})}{x^2} = 1 I have used cos x \cos x ~ 1 x 2 2 ! 1 - \frac{x^2}{2!} when x 0 x \to 0

3.- Use L'Hopital's Rule twice

I have got another :), write it as:- lim x 0 ( ( e x 1 ) x ) 2 × 1 e x = 1 \lim_{x\to 0}\left(\dfrac{(e^x-1)}{x}\right)^2\times\dfrac {1}{e^x}=1 (Using lim x 0 e x 1 x = 1 \lim_{x\to 0} \dfrac{e^x-1}{x}=1 ).

Rishabh Jain - 5 years, 3 months ago

Approach 4: Use sinh ( x ) = e x e x 2 \sinh(x)=\dfrac{e^x-e^{-x}}{2} .

Nihar Mahajan - 5 years, 3 months ago

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Another approach: Note that the original form allows us to use L'Hopital's Rule. Thus, lim x 0 e x + e x 2 x 2 = lim x 0 d ( e x + e x 2 ) d x d x 2 d x = lim x 0 e x e x 2 x = lim x 0 e x + e x 2 = 1 \lim_{x \to 0} \frac{e^x + e^{-x} - 2}{x^2}=\lim_{x \to 0} \frac{\frac{\text{d}(e^x + e^{-x} - 2)}{\text{d}x}}{\frac{\text{d}x^2}{\text{d}x}}=\lim_{x \to 0} \frac{e^x - e^{-x} }{2x}=\lim_{x \to 0} \frac{e^x + e^{-x}}{2}=1

A Former Brilliant Member - 5 years, 3 months ago

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It is incorrect to write d expression d x \dfrac{d \ \text{expression}}{dx} . You should, instead, write d d x ( expression ) \dfrac{d}{dx} (\text{expression}) . Also, d d x ( e x + e x 2 ) = e x e x e x e x 2 \dfrac{d}{dx} (e^x + e^{-x} - 2) = e^x - e^{-x} \neq e^x - e^{-x} - 2 .

This is equivalent to Guillermo Templado's third method.

Jake Lai - 5 years, 3 months ago

To the person who posted the question: please change the current n 0 n \rightarrow 0 to x 0 x \rightarrow 0 .

Shaun Leong - 5 years, 3 months ago

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Thanks , I have edited it. (Its surprising I didn't notice it O.o)

Nihar Mahajan - 5 years, 3 months ago
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Mar 15, 2021

lim x 0 e x + e x 2 x 2 = 1 \displaystyle \lim _ { x \to 0 } \frac { e ^ { x } + e ^ { -x } - 2 } { x ^ { 2 } } = 1 .

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