L'Hospital's Rule? (2)

Calculus Level 3

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

Find the value of L L .

Clarification : e ≈ 2.71828 e \approx 2.71828 is the Euler's number .

e e None of these choices − e 2 -\dfrac{e}{2} e 2 \dfrac{e}{2} − e -e Limit does not exist

Sabhrant Sachan by Sabhrant Sachan , 21, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Sabhrant Sachan
May 30, 2016

Relevant wiki: Maclaurin Series

we know the expansion of ln ⁡ ( x + 1 ) , it is \text{we know the expansion of } \ln{(x+1)} \text{, it is }

ln ⁡ ( x + 1 ) = x − x 2 2 + x 3 3 + ⋯ 1 x ln ⁡ ( x + 1 ) = 1 − x 2 + x 2 3 + ⋯ e 1 x ln ⁡ ( x + 1 ) = e 1 − x 2 + x 2 3 + ⋯ ( x + 1 ) 1 x = e ⋅ e − x 2 + x 2 3 + ⋯ ( x + 1 ) 1 x = e ⋅ ( 1 − 1 2 x + 11 24 x 2 + ⋯ ) ( x + 1 ) 1 x = e − e 2 x + 11 e 24 x 2 + ⋯ Plugging this value in our Limit L = lim ⁡ x → 0 e − e 2 x + 11 e 24 x 2 + ⋯ − e x ⟹ − e 2 \ln{(x+1)} = x-\dfrac{x^2}{2}+\dfrac{x^3}{3}+\cdots \\ \dfrac{1}{x}\ln{(x+1)} = 1-\dfrac{x}{2}+\dfrac{x^2}{3}+\cdots \\ e^{\frac{1}{x}\ln{(x+1)}}=e^{1-\frac{x}{2}+\frac{x^2}{3}+\cdots} \\ (x+1)^{\frac1{x}}=e\cdot e^{-\frac{x}{2}+\frac{x^2}{3}+\cdots} \\ (x+1)^{\frac1{x}}=e\cdot(1-\dfrac{1}{2}x+\dfrac{11}{24}x^2+\cdots) \\ (x+1)^{\frac1{x}}=e-\dfrac{e}{2}x+\dfrac{11e}{24}x^2+\cdots \\ \text{Plugging this value in our Limit } \\ L = \displaystyle\lim_{x \to 0} \dfrac{\cancel{e}-\dfrac{e}{2}x+\dfrac{11e}{24}x^2+\cdots \cancel{-e}}{x} \implies \boxed{-\dfrac{e}{2}}

Alternate Method : For those people who get emotional , when they cannot use L’Hospital’s Rule :P After taking the derivative of numerator and denominator, we get L = lim ⁡ x → 0 d ( ( x + 1 ) 1 x ) d x Let v = ( x + 1 ) 1 x ln ⁡ v = 1 x ln ⁡ ( x + 1 ) d v d x ⋅ 1 v = − ln ⁡ ( x + 1 ) x 2 + 1 x ( x + 1 ) d v d x = ( x + 1 ) 1 x ( 1 x ( x + 1 ) − ln ⁡ ( x + 1 ) x 2 ) L = lim ⁡ x → 0 ( x + 1 ) 1 x ( 1 x ( x + 1 ) − ln ⁡ ( x + 1 ) x 2 ) L = e lim ⁡ x → 0 ( x x 2 ( x + 1 ) − ( x + 1 ) ln ⁡ ( x + 1 ) x 2 ( x + 1 ) ) L = e lim ⁡ x → 0 ( x − ( x + 1 ) ln ⁡ ( x + 1 ) x 2 ) Using LH rule again L = e lim ⁡ x → 0 1 − ln ⁡ ( x + 1 ) − 1 2 x ⟹ L = − e 2 \text{Alternate Method :} \\ \text{For those people who get emotional , when they cannot use L'Hospital's Rule :P} \\ \text{After taking the derivative of numerator and denominator, we get } \\ L = \displaystyle\lim_{x \to 0} \dfrac{d\left( (x+1)^{\frac1{x}} \right)}{dx} \\ \text{Let } v = (x+1)^{\frac1{x}} \\ \ln{v}=\dfrac1{x}\ln{(x+1)} \\ \dfrac{dv}{dx}\cdot \dfrac{1}{v}= -\dfrac{\ln{(x+1)}}{x^2} + \dfrac{1}{x(x+1)} \\ \dfrac{dv}{dx} = (x+1)^{\frac1{x}}\left( \dfrac{1}{x(x+1)}-\dfrac{\ln{(x+1)}}{x^2} \right) \\ L = \displaystyle\lim_{x \to 0} (x+1)^{\frac1{x}}\left( \dfrac{1}{x(x+1)}-\dfrac{\ln{(x+1)}}{x^2} \right) \\ L = e \displaystyle\lim_{x \to 0} \left( \dfrac{x}{x^2(x+1)}-\dfrac{(x+1)\ln{(x+1)}}{x^2(x+1)} \right) \\ L = e \displaystyle\lim_{x \to 0} \left( \dfrac{x-(x+1)\ln{(x+1)}}{x^2} \right) \quad \text{Using LH rule again}\\ L = e \displaystyle\lim_{x \to 0} \dfrac{1-\ln{(x+1)-1}}{2x} \\ \implies\boxed{ L= -\dfrac{e}{2} }

14 Helpful 0 Interesting 1 Brilliant 0 Confused

@Sambhrant Sachan i got emotuonal ..lol...great solution though

Ujjwal Mani Tripathi - 4 years, 3 months ago

In the second last step didn't you miss a ( x + 1 ) (x+1) in the denominator?

Anik Mandal - 4 years, 8 months ago

Log in to reply

He calculated the limit of (x+1) separately which is zero ...

Ashutosh Kapre - 4 years ago

Log in to reply

Edit :sorry for mistake ...Limit of x+1 is 1

Ashutosh Kapre - 4 years ago

Another doubt In the first approach: How did we get e ( 1 − x 2 + 11 x 2 24 + . . . ) \ e (1-\frac{x}{2}+\frac{11x^2}{24}+...) from the previous step? Thanks

Anik Mandal - 4 years, 8 months ago

Log in to reply

I have used the Expansion of e x = 1 + x + x 2 2 ! ⋯ ∞ e^x = 1+x+\dfrac{x^2}{2!}\cdots \infty , but i have only written a few terms which were important in the calculation.

Sabhrant Sachan - 4 years, 8 months ago

Log in to reply

How did we get the terms ( 1 − x 2 + 11 x 2 24 ) (1-\dfrac{x}{2}+\dfrac{11x^2}{24}) ?I understand the expansion part.But after that..Could you explain

Anik Mandal - 4 years, 8 months ago

in third last step,why to use l hopital again,it could be solved directly,but answer changes in that case

sunny chaturvedi - 4 years ago

Log in to reply

@sunny chaturvedi well , we are allowed to use LH rule if it is the case of 0 0 \dfrac{0}{0} or ∞ ∞ \dfrac{\infty}{\infty} . In the third last step , both Denominator and Numerator are equal to 0 0 .

Sabhrant Sachan - 4 years ago

no no,you got me wrong,i mean to say that its our choice to apply L hopital,,,what i am conveying is that in 3rd last step if we dont use l hopital and evaluate the limit by algebric manipulations then answer comes out -e,why is not correct can you explain?

sunny chaturvedi - 4 years ago
Chew-Seong Cheong
May 31, 2016

Relevant wiki: Maclaurin Series

L = lim ⁡ x → 0 ( 1 + x ) 1 x − e x = lim ⁡ x → 0 e ln ⁡ ( 1 + x ) x − e x Using Maclaurin series = lim ⁡ x → 0 e x − x 2 2 + x 3 3 − x 4 4 + . . . x − e x = lim ⁡ x → 0 e 1 − x 2 + x 2 3 − x 3 4 + . . . − e x = lim ⁡ x → 0 e ( e − x 2 + x 2 3 − x 3 4 + . . . − 1 ) x Let y = − x 2 + x 2 3 − x 3 4 + . . . = lim ⁡ x → 0 e ( e y − 1 ) x Using Maclaurin series again = lim ⁡ x → 0 e ( 1 − y + y 2 2 ! − y 3 3 ! + . . . − 1 ) x = lim ⁡ x → 0 − e ( y − y 2 2 ! + y 3 3 ! − . . . ) x = lim ⁡ x → 0 − e ( x 2 − O ( x 2 ) ) x = lim ⁡ x → 0 − e ( 1 2 − O ( x ) ) = − e 2 \begin{aligned} L & = \lim_{x \to 0} \frac{(1+x)^\frac 1x-e}{x} \\ & = \lim_{x \to 0} \frac{e^{\frac{\color{#3D99F6}{\ln(1+x)}}{x}}-e}{x} \quad \quad \small \color{#3D99F6}{\text{Using Maclaurin series}} \\ & = \lim_{x \to 0} \frac{e^{\frac{\color{#3D99F6}{x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...}}{x}}-e}{x} \\ & = \lim_{x \to 0} \frac{e^{1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+...}-e}{x} \\ & = \lim_{x \to 0} \frac{e \big( e^{\color{#3D99F6}{-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+...}}-1\big)}{x} \quad \quad \small \color{#3D99F6}{\text{Let }y = -\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+...} \\ & = \lim_{x \to 0} \frac{e \big(\color{#3D99F6}{e^{y}}-1\big)}{x} \quad \quad \small \color{#3D99F6}{\text{Using Maclaurin series again}} \\ & = \lim_{x \to 0} \frac{e \left(\color{#3D99F6}{1- y + \frac{y^2}{2!}-\frac{y^3}{3!}+...}-1\right)}{x} \\ & = \lim_{x \to 0} \frac{- e \left(y - \frac{y^2}{2!}+\frac{y^3}{3!}-...\right)}{x} \\ & = \lim_{x \to 0} \frac{- e \left(\frac{x}{2} - O(x^2) \right)}{x} \\ & = \lim_{x \to 0} - e \left(\frac{1}{2} - O(x) \right) \\ & = \boxed{ -\dfrac{e}{2}} \end{aligned}

12 Helpful 0 Interesting 1 Brilliant 0 Confused
Swapnil Vatsal
Jun 11, 2017

The expansion of (1+x)^1/x is e(1-x/2+11x^2/24....

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...