L'Hopital's Rule? Think it twice!

Calculus Level 5

L = lim ⁡ x → 0 x 2 sin ⁡ 1 x + 2 x ( 1 + x ) 1 / x − e \large L = \lim_{x\to0} \dfrac{x^2 \sin \frac{1}{x} + 2x}{(1+x)^{1/x} - e}

Find the value of the closed form of L L to 3 decimal places.

Inspiration


The answer is -1.47152.

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Chan Lye Lee by Chan Lye Lee , 44, Singapore

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2 solutions

Chan Lye Lee
May 3, 2017

L = lim ⁡ x → 0 x 2 sin ⁡ 1 x + 2 x x x ( 1 + x ) 1 / x − e = lim ⁡ x → 0 ( x sin ⁡ 1 x + 2 ) ( − 2 e ) = ( 2 ) ( − 2 e ) = − 4 e L =\lim_{x\to0} \dfrac{x^2 \sin \frac{1}{x} + 2x} {x} \dfrac{x}{(1+x)^{1/x} - e} =\lim_{x\to0} \left(x \sin \frac{1}{x} + 2\right) \left(\dfrac{-2}{e}\right) =\left(2\right)\left(\dfrac{-2}{e}\right)=\dfrac{-4}{e}

8 Helpful 0 Interesting 0 Brilliant 0 Confused

Why u wrote Lhostital think twice??. I got the answer using Lhostital very easily.

Nivedit Jain - 3 years, 12 months ago

Ok i got it I am wrong.

Nivedit Jain - 3 years, 12 months ago

how did u do the second part i mean the -2/e part how did it come?

blackjack 699 - 2 years, 7 months ago

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From the Inspiration that I mention.

Chan Lye Lee - 2 years, 7 months ago
Kushal Dey
May 20, 2021

Just to add to Chan Lye Lee's solution (because my approach was also totally the same)
We know, e^x=1+x+x²/2!+...... and ln(1+x)=x-x²/2+x³/3........
(1+x)^(1/x)=e^(ln(1+x)/x)=e^((x-x²/2+x³/3....)/x)=e^(1-x/2+x²/3....)
=e e^(-x/2+x²/3.....)=e (1+(-x/2+x²/3....)+(-x/2+x²/3....)²/2+.....)=e(1-x/2+11/24x²+.....). This is basically the power series expansion of denominator which will help to evaluate the limit.


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