L'Hopital's Rule Might Not Help Here

Calculus Level 5

lim ⁡ x → 0 ( 1 x 2 − e x ( e x − 1 ) 2 ) \lim_{x\to 0} \left(\dfrac{1}{x^2} - \dfrac{e^x}{(e^x-1)^2}\right)

Find the reciprocal of the above limit.


The answer is 12.

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Ariel Gershon by Ariel Gershon , 26, Canada

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3 solutions

Chew-Seong Cheong
Feb 16, 2016

lim ⁡ x → 0 ( 1 x 2 − e x ( e x − 1 ) 2 ) = lim ⁡ x → 0 ( 1 x 2 − e x e 2 x − 2 e x + 1 ) = lim ⁡ x → 0 ( 1 x 2 − 1 e x − 2 + e − x ) By Maclaurin series = lim ⁡ x → 0 ( 1 x 2 − 1 2 ( x 2 2 ! + x 4 4 ! + x 6 6 ! + . . . ) ) = lim ⁡ x → 0 2 ( x 2 2 ! + x 4 4 ! + x 6 6 ! + . . . ) − x 2 2 x 2 ( x 2 2 ! + x 4 4 ! + x 6 6 ! + . . . ) = lim ⁡ x → 0 2 ( x 4 4 ! + x 6 6 ! + x 8 8 ! + . . . ) 2 ( x 4 2 ! + x 6 4 ! + x 8 8 ! + . . . ) = lim ⁡ x → 0 1 4 ! + x 2 6 ! + x 4 8 ! + . . . 1 2 ! + x 2 4 ! + x 4 8 ! + . . . = 1 4 ! 1 2 ! = 2 ! 4 ! = 1 12 \begin{aligned} \lim_{x \to 0} \left(\frac{1}{x^2} - \frac{e^x}{(e^x-1)^2}\right) & = \lim_{x \to 0} \left(\frac{1}{x^2} - \frac{e^x}{e^{2x} -2e^x +1}\right) \\ & = \lim_{x \to 0} \left(\frac{1}{x^2} - \frac{1}{\color{#3D99F6}{e^x -2 + e^{-x}}}\right) \quad \quad \small \color{#3D99F6}{\text{By Maclaurin series}} \\ & = \lim_{x \to 0} \left(\frac{1}{x^2} - \frac{1}{\color{#3D99F6}{2\left(\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+...\right)}} \right) \\ & = \lim_{x \to 0} \frac{2\left(\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+... \right)- x^2}{2x^2\left(\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+...\right)} \\ & = \lim_{x \to 0} \frac{2\left(\frac{x^4}{4!}+\frac{x^6}{6!}+\frac{x^8}{8!}+... \right)}{2\left(\frac{x^4}{2!}+\frac{x^6}{4!}+\frac{x^8}{8!}+...\right)} \\ & = \lim_{x \to 0} \frac{\frac{1}{4!}+\frac{x^2}{6!}+\frac{x^4}{8!}+... }{\frac{1}{2!}+\frac{x^2}{4!}+\frac{x^4}{8!}+...} \\ & = \frac{\frac{1}{4!}}{\frac{1}{2!}} = \frac{2!}{4!} = \frac{1}{\boxed{12}} \end{aligned}

14 Helpful 0 Interesting 1 Brilliant 0 Confused

Interesting approach! You made one mistake though: the Maclaurin series of e x − 2 + e − x e^x - 2 + e^{-x} is 2 ( x 2 2 ! + x 4 4 ! + x 6 6 ! + . . . ) 2\left(\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+...\right) . So actually, the step before you use L'Hopital's rule would look like this: lim ⁡ x → 0 2 ( x 4 4 ! + x 6 6 ! + . . . ) 2 ( x 4 2 ! + x 6 4 ! + . . . ) \lim_{x\to 0} \dfrac{2\left(\frac{x^4}{4!}+\frac{x^6}{6!}+...\right)}{2\left(\frac{x^4}{2!}+\frac{x^6}{4!}+...\right)} Looks like you don't even need L'Hopital's rule in this case!

Ariel Gershon - 5 years, 3 months ago

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Thanks. I will amend the solution. Nice problem.

Chew-Seong Cheong - 5 years, 3 months ago

So finally a question to which you couldn't (and didn't) apply L'Hopital's Rule.

Anyway, exceptionally brilliant solution as always. Cheers! (+1)

Arkajyoti Banerjee - 4 years, 4 months ago

Elegent approach and I really love doing with this.

Naren Bhandari - 3 years, 8 months ago

=limx>0 e 2 x + 1 − 2 e x − e x ∗ x 2 x 2 ( e x − 1 ) 2 \frac{e^{2x}+1-2e^{x}-e^{x}*x^{2}}{x^{2}(e^{x}-1)^{2}} =limx>0 e 2 x + 1 − 2 e x − e x ∗ x 2 x 4 ( e x − 1 ) 2 x 2 \frac{e^{2x}+1-2e^{x}-e^{x}*x^{2}}{x^{4}\frac{(e^{x}-1)^{2}}{x^{2}}} =limx>0 e 2 x + 1 − 2 e x − e x ∗ x 2 x 4 \frac{e^{2x}+1-2e^{x}-e^{x}*x^{2}}{x^{4}} Now if the limit has to exist we can just find the coefficient of x 4 x^{4} in numerator by applying the expansion of e. = 1 + 2 x + 4 x 2 2 ! + 8 x 3 3 ! + 16 x 4 4 ! . . . + 1 − 2 − 2 x − 2 x 2 2 ! − 2 x 3 3 ! − 2 x 4 4 ! . . . . − x 2 ( 1 + x + x 2 2 ! . . . ) 1+2x+\frac{4x^{2}}{2!}+\frac{8x^{3}}{3!}+\frac{16x^{4}}{4!}...+1-2-2x-\frac{2x^{2}}{2!}-\frac{2x^{3}}{3!}-\frac{2x^{4}}{4!}....-x^{2}(1+x+\frac{x^{2}}{2!}...) Taking out the coefficient of x 4 x^{4}

16 4 ! − 2 4 ! − 1 2 ! \frac{16}{4!}-\frac{2}{4!}-\frac{1}{2!} = 1 12 \frac{1}{12}

We can be sure that terms having power less than 4 will cancel out and more than 4 will become 0

3 Helpful 0 Interesting 0 Brilliant 0 Confused

the LaTeX code for the limit of a function f ( x ) f(x) is

\lim _{ x\rightarrow a }{ f(x) }

lim ⁡ x → a f ( x ) \lim _{ x\rightarrow a }{ f(x) } (what the math would look like)

or you can add \displaystyle so it'll look like this

lim ⁡ x → a f ( x ) \displaystyle\lim _{ x\rightarrow a }{ f(x) }

Hamza A - 5 years, 3 months ago

lim ⁡ x → 0 ( e x − 1 ) 2 − x 2 e x ( x ( e x − 1 ) ) 2 \large \displaystyle \lim_{x \to 0} \frac{\left(e^x-1\right)^2-x^2e^x}{\left(x\left(e^x-1\right)\right)^2}

Plugging in e x = 1 + x + x 2 2 ! + x 3 3 ! + . . . e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+... (Taylor's Expansion for e x e^x ):

= lim ⁡ x → 0 ( 1 + x + x 2 2 ! + x 3 3 ! + . . . − 1 ) 2 − x 2 ( 1 + x + x 2 2 ! + x 3 3 ! + . . . ) ( x ( 1 + x + x 2 2 ! + x 3 3 ! + . . . − 1 ) ) 2 =\large \displaystyle \lim_{x \to 0} \frac{\left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...-1\right)^2-x^2\left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...\right)}{\left(x\left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...-1\right)\right)^2}

= lim ⁡ x → 0 ( x + x 2 2 ! + x 3 3 ! + . . . ) 2 − x 2 ( 1 + x + x 2 2 ! + x 3 3 ! + . . . ) ( x ( x + x 2 2 ! + x 3 3 ! + . . . ) ) 2 =\large \displaystyle \lim_{x \to 0} \frac{\left(x+\frac{x^2}{2!}+\frac{x^3}{3!}+...\right)^2-x^2\left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...\right)}{\left(x\left(x+\frac{x^2}{2!}+\frac{x^3}{3!}+...\right)\right)^2}

= lim ⁡ x → 0 ( x + x 2 2 ! + x 3 3 ! + . . . ) ( x + x 2 2 ! + x 3 3 ! + . . . ) − x 2 − x 3 − x 4 2 ! − x 5 3 ! − . . . ( x 2 + x 3 2 ! + x 4 3 ! + . . . ) 2 =\large \displaystyle \lim_{x \to 0} \frac{\left(x+\frac{x^2}{2!}+\frac{x^3}{3!}+...\right)\left(x+\frac{x^2}{2!}+\frac{x^3}{3!}+...\right)-x^2-x^3-\frac{x^4}{2!}-\frac{x^5}{3!}-...}{\left(x^2+\frac{x^3}{2!}+\frac{x^4}{3!}+...\right)^2}

From here, it is pretty clear that the lowest powered term in the denominator will be x 4 x^4 . Evidently, for limit to exist and be non infinite, terms with a power lesser than 4 4 in the numerator must cancel out and the terms larger than 4 4 must equate to 0 0 as x → 0 x \to 0 . Hence, we collect all the x 4 x^4 terms in the numerator to prepare for cancellation.

= lim ⁡ x → 0 ( x ⋅ x 3 3 ! + x 2 2 ! ⋅ x 2 2 ! + x 3 3 ! ⋅ x − x 4 2 ! ) x 4 =\large \displaystyle \lim_{x \to 0} \frac{\left(x\cdot \frac{x^3}{3!}+\frac{x^2}{2!}\cdot \frac{x^2}{2!}+\frac{x^3}{3!}\cdot x-\frac{x^4}{2!}\right)}{x^4}

= 1 6 + 1 4 + 1 6 − 1 2 = 1 12 =\frac{1}{6}+\frac{1}{4}+\frac{1}{6}-\frac{1}{2} = \boxed{\frac{1}{12}}

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