Evaluate $limlimits_{x to 1}dfrac{x-x^x}{1-x+ln x}$.

2

$begingroup$

Problem

Evaluate $$limlimits_{x to 1}frac{x-x^x}{1-x+ln x}$$.

Solution

Consider using Taylor's formula. Expand $x^x $ and $ln x$ at $x=1$. We obtain

$$x^x=1+(x-1)+(x-1)^2+o((x-1)^2);$$
$$ln x=(x-1)-frac{1}{2}(x-1)^2+o((x-1)^2).$$
Therefore
begin{align*}
lim_{x to 1}
frac{x-x^x}{1-x+ln x}&=lim_{x to 1}frac{x-[1+(x-1)+(x-1)^2+o((x-1)^2)]}{1-x+[(x-1)-frac{1}{2}(x-1)^2+o((x-1)^2)]}\
&=lim_{x to 1}frac{-(x-1)^2-o((x-1)^2)}{-frac{1}{2}(x-1)^2+o((x-1)^2)}\
&=2.
end{align*}

Please correct me if I'm wrong! Can we solve it by L'Hospital's rule?

real-analysis limits proof-verification

share|cite|improve this question

edited Jan 16 at 14:42

Michael Rozenberg

104k1892197

asked Jan 16 at 14:23

mengdie1982mengdie1982

4,897618

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do you get the expansion of $x^x$ ?
  $endgroup$
  – Yves Daoust
  Jan 16 at 14:40
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  l'Hospital's rule can be used because it is the case $0/0$ and the denominator derivative $(x-1+ln x)'neq 0$ if $xneq 1.$
  $endgroup$
  – user376343
  Jan 16 at 16:15
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Problem

Evaluate $$limlimits_{x to 1}frac{x-x^x}{1-x+ln x}$$.

Solution

Consider using Taylor's formula. Expand $x^x $ and $ln x$ at $x=1$. We obtain

$$x^x=1+(x-1)+(x-1)^2+o((x-1)^2);$$
$$ln x=(x-1)-frac{1}{2}(x-1)^2+o((x-1)^2).$$
Therefore
begin{align*}
lim_{x to 1}
frac{x-x^x}{1-x+ln x}&=lim_{x to 1}frac{x-[1+(x-1)+(x-1)^2+o((x-1)^2)]}{1-x+[(x-1)-frac{1}{2}(x-1)^2+o((x-1)^2)]}\
&=lim_{x to 1}frac{-(x-1)^2-o((x-1)^2)}{-frac{1}{2}(x-1)^2+o((x-1)^2)}\
&=2.
end{align*}

Please correct me if I'm wrong! Can we solve it by L'Hospital's rule?

real-analysis limits proof-verification

share|cite|improve this question

edited Jan 16 at 14:42

Michael Rozenberg

104k1892197

asked Jan 16 at 14:23

mengdie1982mengdie1982

4,897618

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do you get the expansion of $x^x$ ?
  $endgroup$
  – Yves Daoust
  Jan 16 at 14:40
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  l'Hospital's rule can be used because it is the case $0/0$ and the denominator derivative $(x-1+ln x)'neq 0$ if $xneq 1.$
  $endgroup$
  – user376343
  Jan 16 at 16:15
  
  
  

  
  
  
  

add a comment | 

2

2

2

$begingroup$

Problem

Evaluate $$limlimits_{x to 1}frac{x-x^x}{1-x+ln x}$$.

Solution

Consider using Taylor's formula. Expand $x^x $ and $ln x$ at $x=1$. We obtain

$$x^x=1+(x-1)+(x-1)^2+o((x-1)^2);$$
$$ln x=(x-1)-frac{1}{2}(x-1)^2+o((x-1)^2).$$
Therefore
begin{align*}
lim_{x to 1}
frac{x-x^x}{1-x+ln x}&=lim_{x to 1}frac{x-[1+(x-1)+(x-1)^2+o((x-1)^2)]}{1-x+[(x-1)-frac{1}{2}(x-1)^2+o((x-1)^2)]}\
&=lim_{x to 1}frac{-(x-1)^2-o((x-1)^2)}{-frac{1}{2}(x-1)^2+o((x-1)^2)}\
&=2.
end{align*}

Please correct me if I'm wrong! Can we solve it by L'Hospital's rule?

real-analysis limits proof-verification

share|cite|improve this question

edited Jan 16 at 14:42

Michael Rozenberg

104k1892197

asked Jan 16 at 14:23

mengdie1982mengdie1982

4,897618

$endgroup$

Problem

Evaluate $$limlimits_{x to 1}frac{x-x^x}{1-x+ln x}$$.

Solution

Consider using Taylor's formula. Expand $x^x $ and $ln x$ at $x=1$. We obtain

$$x^x=1+(x-1)+(x-1)^2+o((x-1)^2);$$
$$ln x=(x-1)-frac{1}{2}(x-1)^2+o((x-1)^2).$$
Therefore
begin{align*}
lim_{x to 1}
frac{x-x^x}{1-x+ln x}&=lim_{x to 1}frac{x-[1+(x-1)+(x-1)^2+o((x-1)^2)]}{1-x+[(x-1)-frac{1}{2}(x-1)^2+o((x-1)^2)]}\
&=lim_{x to 1}frac{-(x-1)^2-o((x-1)^2)}{-frac{1}{2}(x-1)^2+o((x-1)^2)}\
&=2.
end{align*}

Please correct me if I'm wrong! Can we solve it by L'Hospital's rule?

real-analysis limits proof-verification

real-analysis limits proof-verification

share|cite|improve this question

edited Jan 16 at 14:42

Michael Rozenberg

104k1892197

asked Jan 16 at 14:23

mengdie1982mengdie1982

4,897618

share|cite|improve this question

edited Jan 16 at 14:42

Michael Rozenberg

104k1892197

asked Jan 16 at 14:23

mengdie1982mengdie1982

4,897618

share|cite|improve this question

share|cite|improve this question

edited Jan 16 at 14:42

Michael Rozenberg

104k1892197

edited Jan 16 at 14:42

Michael Rozenberg

104k1892197

edited Jan 16 at 14:42

Michael Rozenberg

104k1892197

104k1892197

asked Jan 16 at 14:23

mengdie1982mengdie1982

4,897618

asked Jan 16 at 14:23

mengdie1982mengdie1982

4,897618

asked Jan 16 at 14:23

mengdie1982mengdie1982

4,897618

4,897618

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do you get the expansion of $x^x$ ?
  $endgroup$
  – Yves Daoust
  Jan 16 at 14:40
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  l'Hospital's rule can be used because it is the case $0/0$ and the denominator derivative $(x-1+ln x)'neq 0$ if $xneq 1.$
  $endgroup$
  – user376343
  Jan 16 at 16:15
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do you get the expansion of $x^x$ ?
  $endgroup$
  – Yves Daoust
  Jan 16 at 14:40
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  l'Hospital's rule can be used because it is the case $0/0$ and the denominator derivative $(x-1+ln x)'neq 0$ if $xneq 1.$
  $endgroup$
  – user376343
  Jan 16 at 16:15
  
  
  

  
  
  
  

$begingroup$
How do you get the expansion of $x^x$ ?
$endgroup$
– Yves Daoust
Jan 16 at 14:40

$begingroup$
How do you get the expansion of $x^x$ ?
$endgroup$
– Yves Daoust
Jan 16 at 14:40

$begingroup$
l'Hospital's rule can be used because it is the case $0/0$ and the denominator derivative $(x-1+ln x)'neq 0$ if $xneq 1.$
$endgroup$
– user376343
Jan 16 at 16:15

$begingroup$
l'Hospital's rule can be used because it is the case $0/0$ and the denominator derivative $(x-1+ln x)'neq 0$ if $xneq 1.$
$endgroup$
– user376343
Jan 16 at 16:15

add a comment | 

3 Answers
3

active

oldest

votes

2

$begingroup$

$1-x=himplies$

$$lim_{hto0}(1-h)cdotdfrac{1-(1-h)^{-h}}{h+ln(1-h)}=lim_{hto0}dfrac{1-left(1+(-h)(-h)+dfrac{(-h)(-h-1)}2cdot h^2+O(h^3)right)}{h-left(h+dfrac{h^2}2+O(h^3)right)}=dfrac1{dfrac12}$$

So, you are correct

share|cite|improve this answer

answered Jan 16 at 14:32

lab bhattacharjeelab bhattacharjee

226k15157275

$endgroup$

add a comment | 

2

$begingroup$

Why no?
$$lim_{xrightarrow1}frac{x-x^x}{1-x+ln{x}}=lim_{xrightarrow1}frac{1-x^x(1+ln{x})}{-1+frac{1}{x}}=lim_{xrightarrow1}frac{-x^x(1+ln{x})^2-x^xcdotfrac{1}{x}}{-frac{1}{x^2}}=2$$

share|cite|improve this answer

answered Jan 16 at 14:40

Michael RozenbergMichael Rozenberg

104k1892197

$endgroup$

add a comment | 

1

$begingroup$

Another solution by L'Hospital's rule

Let $x=:1+h$. Then
begin{align*}
lim_{x to 1}
frac{x-x^x}{1-x+ln x}&=lim_{h to 0}frac{(1+h)[1-(1+h)^{h}]}{ln(1+h)-h}\
&=lim_{h to 0}frac{e^{hln(1+h)}-1}{h-ln(1+h)}\
&=lim_{h to 0}frac{hln(1+h)}{h-ln(1+h)}\
&=lim_{h to 0}frac{h^2}{h-ln(1+h)}\
&=lim_{h to 0}frac{2h}{1-frac{1}{1+h}}\
&=2lim_{h to 0}(1+h)\
&=2.
end{align*}

share|cite|improve this answer

answered Jan 16 at 14:57

mengdie1982mengdie1982

4,897618

$endgroup$

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

2

$begingroup$

$1-x=himplies$

$$lim_{hto0}(1-h)cdotdfrac{1-(1-h)^{-h}}{h+ln(1-h)}=lim_{hto0}dfrac{1-left(1+(-h)(-h)+dfrac{(-h)(-h-1)}2cdot h^2+O(h^3)right)}{h-left(h+dfrac{h^2}2+O(h^3)right)}=dfrac1{dfrac12}$$

So, you are correct

share|cite|improve this answer

answered Jan 16 at 14:32

lab bhattacharjeelab bhattacharjee

226k15157275

$endgroup$

add a comment | 

2

$begingroup$

$1-x=himplies$

$$lim_{hto0}(1-h)cdotdfrac{1-(1-h)^{-h}}{h+ln(1-h)}=lim_{hto0}dfrac{1-left(1+(-h)(-h)+dfrac{(-h)(-h-1)}2cdot h^2+O(h^3)right)}{h-left(h+dfrac{h^2}2+O(h^3)right)}=dfrac1{dfrac12}$$

So, you are correct

share|cite|improve this answer

answered Jan 16 at 14:32

lab bhattacharjeelab bhattacharjee

226k15157275

$endgroup$

add a comment | 

2

2

2

$begingroup$

$1-x=himplies$

$$lim_{hto0}(1-h)cdotdfrac{1-(1-h)^{-h}}{h+ln(1-h)}=lim_{hto0}dfrac{1-left(1+(-h)(-h)+dfrac{(-h)(-h-1)}2cdot h^2+O(h^3)right)}{h-left(h+dfrac{h^2}2+O(h^3)right)}=dfrac1{dfrac12}$$

So, you are correct

share|cite|improve this answer

answered Jan 16 at 14:32

lab bhattacharjeelab bhattacharjee

226k15157275

$endgroup$

$1-x=himplies$

$$lim_{hto0}(1-h)cdotdfrac{1-(1-h)^{-h}}{h+ln(1-h)}=lim_{hto0}dfrac{1-left(1+(-h)(-h)+dfrac{(-h)(-h-1)}2cdot h^2+O(h^3)right)}{h-left(h+dfrac{h^2}2+O(h^3)right)}=dfrac1{dfrac12}$$

So, you are correct

share|cite|improve this answer

answered Jan 16 at 14:32

lab bhattacharjeelab bhattacharjee

226k15157275

share|cite|improve this answer

share|cite|improve this answer

answered Jan 16 at 14:32

lab bhattacharjeelab bhattacharjee

226k15157275

answered Jan 16 at 14:32

lab bhattacharjeelab bhattacharjee

226k15157275

answered Jan 16 at 14:32

lab bhattacharjeelab bhattacharjee

226k15157275

226k15157275

add a comment | 

add a comment | 

2

$begingroup$

Why no?
$$lim_{xrightarrow1}frac{x-x^x}{1-x+ln{x}}=lim_{xrightarrow1}frac{1-x^x(1+ln{x})}{-1+frac{1}{x}}=lim_{xrightarrow1}frac{-x^x(1+ln{x})^2-x^xcdotfrac{1}{x}}{-frac{1}{x^2}}=2$$

share|cite|improve this answer

answered Jan 16 at 14:40

Michael RozenbergMichael Rozenberg

104k1892197

$endgroup$

add a comment | 

2

$begingroup$

Why no?
$$lim_{xrightarrow1}frac{x-x^x}{1-x+ln{x}}=lim_{xrightarrow1}frac{1-x^x(1+ln{x})}{-1+frac{1}{x}}=lim_{xrightarrow1}frac{-x^x(1+ln{x})^2-x^xcdotfrac{1}{x}}{-frac{1}{x^2}}=2$$

share|cite|improve this answer

answered Jan 16 at 14:40

Michael RozenbergMichael Rozenberg

104k1892197

$endgroup$

add a comment | 

2

2

2

$begingroup$

Why no?
$$lim_{xrightarrow1}frac{x-x^x}{1-x+ln{x}}=lim_{xrightarrow1}frac{1-x^x(1+ln{x})}{-1+frac{1}{x}}=lim_{xrightarrow1}frac{-x^x(1+ln{x})^2-x^xcdotfrac{1}{x}}{-frac{1}{x^2}}=2$$

share|cite|improve this answer

answered Jan 16 at 14:40

Michael RozenbergMichael Rozenberg

104k1892197

$endgroup$

Why no?
$$lim_{xrightarrow1}frac{x-x^x}{1-x+ln{x}}=lim_{xrightarrow1}frac{1-x^x(1+ln{x})}{-1+frac{1}{x}}=lim_{xrightarrow1}frac{-x^x(1+ln{x})^2-x^xcdotfrac{1}{x}}{-frac{1}{x^2}}=2$$

share|cite|improve this answer

answered Jan 16 at 14:40

Michael RozenbergMichael Rozenberg

104k1892197

share|cite|improve this answer

share|cite|improve this answer

answered Jan 16 at 14:40

Michael RozenbergMichael Rozenberg

104k1892197

answered Jan 16 at 14:40

Michael RozenbergMichael Rozenberg

104k1892197

answered Jan 16 at 14:40

Michael RozenbergMichael Rozenberg

104k1892197

104k1892197

add a comment | 

add a comment | 

1

$begingroup$

Another solution by L'Hospital's rule

Let $x=:1+h$. Then
begin{align*}
lim_{x to 1}
frac{x-x^x}{1-x+ln x}&=lim_{h to 0}frac{(1+h)[1-(1+h)^{h}]}{ln(1+h)-h}\
&=lim_{h to 0}frac{e^{hln(1+h)}-1}{h-ln(1+h)}\
&=lim_{h to 0}frac{hln(1+h)}{h-ln(1+h)}\
&=lim_{h to 0}frac{h^2}{h-ln(1+h)}\
&=lim_{h to 0}frac{2h}{1-frac{1}{1+h}}\
&=2lim_{h to 0}(1+h)\
&=2.
end{align*}

share|cite|improve this answer

answered Jan 16 at 14:57

mengdie1982mengdie1982

4,897618

$endgroup$

add a comment | 

1

$begingroup$

Another solution by L'Hospital's rule

Let $x=:1+h$. Then
begin{align*}
lim_{x to 1}
frac{x-x^x}{1-x+ln x}&=lim_{h to 0}frac{(1+h)[1-(1+h)^{h}]}{ln(1+h)-h}\
&=lim_{h to 0}frac{e^{hln(1+h)}-1}{h-ln(1+h)}\
&=lim_{h to 0}frac{hln(1+h)}{h-ln(1+h)}\
&=lim_{h to 0}frac{h^2}{h-ln(1+h)}\
&=lim_{h to 0}frac{2h}{1-frac{1}{1+h}}\
&=2lim_{h to 0}(1+h)\
&=2.
end{align*}

share|cite|improve this answer

answered Jan 16 at 14:57

mengdie1982mengdie1982

4,897618

$endgroup$

add a comment | 

1

1

1

$begingroup$

Another solution by L'Hospital's rule

Let $x=:1+h$. Then
begin{align*}
lim_{x to 1}
frac{x-x^x}{1-x+ln x}&=lim_{h to 0}frac{(1+h)[1-(1+h)^{h}]}{ln(1+h)-h}\
&=lim_{h to 0}frac{e^{hln(1+h)}-1}{h-ln(1+h)}\
&=lim_{h to 0}frac{hln(1+h)}{h-ln(1+h)}\
&=lim_{h to 0}frac{h^2}{h-ln(1+h)}\
&=lim_{h to 0}frac{2h}{1-frac{1}{1+h}}\
&=2lim_{h to 0}(1+h)\
&=2.
end{align*}

share|cite|improve this answer

answered Jan 16 at 14:57

mengdie1982mengdie1982

4,897618

$endgroup$

Another solution by L'Hospital's rule

Let $x=:1+h$. Then
begin{align*}
lim_{x to 1}
frac{x-x^x}{1-x+ln x}&=lim_{h to 0}frac{(1+h)[1-(1+h)^{h}]}{ln(1+h)-h}\
&=lim_{h to 0}frac{e^{hln(1+h)}-1}{h-ln(1+h)}\
&=lim_{h to 0}frac{hln(1+h)}{h-ln(1+h)}\
&=lim_{h to 0}frac{h^2}{h-ln(1+h)}\
&=lim_{h to 0}frac{2h}{1-frac{1}{1+h}}\
&=2lim_{h to 0}(1+h)\
&=2.
end{align*}

share|cite|improve this answer

answered Jan 16 at 14:57

mengdie1982mengdie1982

4,897618

share|cite|improve this answer

share|cite|improve this answer

answered Jan 16 at 14:57

mengdie1982mengdie1982

4,897618

answered Jan 16 at 14:57

mengdie1982mengdie1982

4,897618

answered Jan 16 at 14:57

mengdie1982mengdie1982

4,897618

4,897618

add a comment | 

add a comment | 

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