Reciprocal Of Logarithm Minus Reciprocal

Calculus Level 3

lim ⁡ a → 1 ( 1 ln ⁡ a − 1 a − 1 ) \large \lim_{a\to 1} \left(\dfrac{1}{\ln a} - \dfrac{1}{a-1}\right)

Find the value of the closed form of the above limit.


The answer is 0.5.

View wiki
Ariel Gershon by Ariel Gershon , 26, Canada

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

Chew-Seong Cheong
Oct 18, 2016

Relevant wiki: L'Hopital's Rule - Basic

L = lim ⁡ a → 1 ( 1 ln ⁡ a − 1 a − 1 ) = lim ⁡ a → 1 a − 1 − ln ⁡ a ln ⁡ a ( a − 1 ) This is a 0/0 case and L’H o ˆ pital’s rule applies. = lim ⁡ a → 1 1 − 1 a a − 1 a + ln ⁡ a Differentiate up and down w.r.t x = lim ⁡ a → 1 a − 1 a − 1 + a ln ⁡ a Again 0/0 case. = lim ⁡ a → 1 1 1 + a a + ln ⁡ a Differentiate up and down w.r.t x = 1 2 = 0.5 \begin{aligned} L & = \lim_{a \to 1} \left(\frac 1{\ln a} - \frac 1{a-1} \right) \\ & = \lim_{a \to 1} \frac {a-1-\ln a}{\ln a(a-1)} & \small \color{#3D99F6}{\text{This is a 0/0 case and L'Hôpital's rule applies.}} \\ & = \lim_{a \to 1} \frac {1-\frac 1a}{\frac {a-1}a + \ln a} & \small \color{#3D99F6}{\text{Differentiate up and down w.r.t }x} \\ & = \lim_{a \to 1} \frac {a-1}{a-1 + a\ln a} & \small \color{#3D99F6}{\text{Again 0/0 case.}} \\ & = \lim_{a \to 1} \frac 1{1 + \frac aa + \ln a} & \small \color{#3D99F6}{\text{Differentiate up and down w.r.t }x} \\ & = \frac 12 = \boxed{0.5} \end{aligned}

17 Helpful 0 Interesting 0 Brilliant 0 Confused

+1 Your solution is a great read:

  1. Thanks for adding in the reasoning at each step. That makes it very easy to follow what you're doing.

Thanks for contributing and helping other members aspire to be like you!

Calvin Lin Staff - 4 years, 8 months ago
Viki Zeta
Oct 17, 2016

L = lim ⁡ a → 1 ( 1 ln ⁡ ( a ) − 1 a − 1 ) = lim ⁡ a → 1 a − 1 − ln ⁡ ( a ) ln ⁡ ( a ) ( a − 1 ) Putting a = 1, we get an indeterminate form 0 0 So, by L’Hospital’s rule L = lim ⁡ a → 1 a − 1 a 1 − 1 a + ln ⁡ ( a ) Even now we have an indeterminate form 0 0 Again using L’Hospital’s rule L = lim ⁡ a → 1 1 a 2 a + 1 a 2 = 1 a + 1 Now substitute a = 1 L = 1 1 + 1 = 1 2 = 0.5 L = \lim_{a\rightarrow 1} (\dfrac{1}{\ln(a)} - \dfrac{1}{a-1}) = \lim_{a\rightarrow 1} \dfrac{a-1-\ln(a)}{\ln(a)(a-1)}\\ \text{Putting a = 1, we get an indeterminate form }\dfrac{0}{0} \\ \text{So, by L'Hospital's rule} \\ L = \lim_{a\rightarrow 1} \dfrac{\dfrac{a-1}{a}}{1-\dfrac{1}{a}+\ln(a)} \\ \text{Even now we have an indeterminate form }\dfrac{0}{0} \\ \text{Again using L'Hospital's rule} \\ L = \lim_{a\rightarrow 1} \dfrac{\dfrac{1}{a^2}}{\dfrac{a+1}{a^2}} = \dfrac{1}{a+1}\\ \text{Now substitute a = 1} \\ \boxed{L = \dfrac{1}{1+1} = \dfrac{1}{2} = 0.5}

9 Helpful 0 Interesting 0 Brilliant 0 Confused

This is a good start. However, your explanation has an error.

  1. If the indeterminate form is 1 0 \frac{1}{0} , then we cannot apply l'hopital's rule .

Your solution has a simple fix. Do you see it?

Calvin Lin Staff - 4 years, 8 months ago

Log in to reply

Oh yeah, it's not 1/0. Thanks I'll edit when am in my PC

Viki Zeta - 4 years, 8 months ago
Prakhar Bindal
Oct 18, 2016

Firstly let a = h+1

Our limit becomes 1/ln(h+1) - 1/h

Use Maclaurin's series for ln(1+x) now

Limit becomes

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Can you fill in more details? How does using the Maclaurin series help directly? Seems like there are a few more steps to complete the solution.

Calvin Lin Staff - 4 years, 8 months ago

If you think about it, in order to figure out the McLaurin series, you first have to find the value of this limit, not the other way around..

Ariel Gershon - 4 years, 7 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...