Pls help this beginner: Limit x->0+ [(x+e^x)^(1/x)]

[Revan]

I didn't solve it pls help me

Problem:

Limit x->0+ [(x+e^x)^(1/x)]

Tanks for your interest

 

[Revan]

And one more question which solition are pratical,limit or derivation?

my eng. is not good sorry

 

[HallsofIvy]

I'm not sure what you mean by "limit or derivation". The problem asks you to find the limit and doing what the problem asks is always "practical"!

The first thing I would do is get rid of that exponent by taking the logarithm:
If \(\displaystyle y= (1+ e^x)^{1/x}\) then \(\displaystyle ln(y)= \frac{1}{x}(1+ e^x)\) so \(\displaystyle x ln(y)= 1+ e^x\).

Now the limit on the right side is 1+ 1= 2 while on the left the "x" term goes to 0. If the limit of ln(y) were finite, the limit on the left would be 0, not 2. So the limit of ln(y) cannot be any finite number. That doesn't necessarily mean that y does not have a finite limit. If the limit of ln(y) were \(\displaystyle -\infty\), the limit of y would be 0. As a check, when x= 0.1, \(\displaystyle (1+ e^x)^{1/x}= (1+ e^{0.1})^{10}= 1709.5...\).

No, that's not anywhere near 0 so the limit is not 0. The only other alternative is that lim y is infinity (does not exist).


THIS IS WRONG! SEE BELOW.

 

[ahorn]

I'm not sure what you mean by "limit or derivation". The problem asks you to find the limit and doing what the problem asks is always "practical"!

The first thing I would do is get rid of that exponent by taking the logarithm:
If \(\displaystyle y= (1+ e^x)^{1/x}\) then \(\displaystyle ln(y)= \frac{1}{x}(1+ e^x)\) so \(\displaystyle x ln(y)= 1+ e^x\).

Now the limit on the right side is 1+ 1= 2 while on the left the "x" term goes to 0. If the limit of ln(y) were finite, the limit on the left would be 0, not 2. So the limit of ln(y) cannot be any finite number. That doesn't necessarily mean that y does not have a finite limit. If the limit of ln(y) were \(\displaystyle -\infty\), the limit of y would be 0. As a check, when x= 0.1, \(\displaystyle (1+ e^x)^{1/x}= (1+ e^{0.1})^{10}= 1709.5...\).

No, that's not anywhere near 0 so the limit is not 0. The only other alternative is that lim y is infinity (does not exist).

I'm not sure why you substituted 1 for x in the binomial.





I got


\(\displaystyle \log_{x+e^x}y=\frac{1}{x}\\


\frac{ln(y)}{ln(x+e^x)}=\frac{1}{x}\\


ln(y)=\frac{1}{x}ln(x+e^x)\\


xln(y)=ln(x+e^x)\)


using your method.


Taking the lim_x->0+ on both sides, I get 0=ln1

Perhaps that means \(\displaystyle y \in \Re \)?

 

[Deleted member 4993]

I'm not sure why you substituted 1 for x in the binomial.





I got


\(\displaystyle \log_{x+e^x}y=\frac{1}{x}\\


\frac{ln(y)}{ln(x+e^x)}=\frac{1}{x}\\


ln(y)=\frac{1}{x}ln(x+e^x)\\


xln(y)=ln(x+e^x)\)


using your method.


Taking the lim_x->0+ on both sides, I get 0=ln1

Perhaps that means \(\displaystyle y \in \Re \)?

If you plot the function - you would see that y → +∞ for x → 0+

 

[lookagain]

The first thing I would do is get rid of that exponent by taking the logarithm:

If \(\displaystyle \ y= (1+ e^x)^{1/x}\) then \(\displaystyle ln(y)= \frac{1}{x}(1+ e^x)\) so \(\displaystyle x ln(y)= 1+ e^x\).

That is an accident/typo typed above. \(\displaystyle \ \ It \ \ should \ \ be \ \ if \ \ y \ \ = \ (x + e^x)^{1/x}.\)

 

[HallsofIvy]

Not to mention I forgot the "ln" on the right! Starting again:
If \(\displaystyle y= (x+ e^x)^{1/x}\) then \(\displaystyle ln(y)= \frac{ln(x+ e^x)}{x}\). Now the numerator and denominator both go to 0 so we can use L'Hopital's rule. The derivative of \(\displaystyle ln(x+ e^x)\) is \(\displaystyle \frac{1}{x+ e^x}(1+ e^x)\) and the derivative of the denominator is 1. Knowing the limit of ln(y), and the fact that logarithm is a continuous function, lets you find the limit of y.


Thanks to all who pointed out my many errors!

 

[ahorn]

Is the answer e^2?

 

[Deleted member 4993]

Is the answer e^2? Correct

You can verify this with spreadsheet like MS_excel.

[TABLE="width: 256"]
  [TR]
   [TD="class: xl65, width: 64, align: right"]1
[/TD]
   [TD="class: xl65, width: 64, align: right"][COLOR=#ffffff]..........[/COLOR]3.718282
[/TD]
   [TD="class: xl65, width: 64, align: right"][COLOR=#ffffff]..................[/COLOR]-1
[/TD]
   [TD="class: xl65, width: 64, align: right"]-1.58198
[/TD]
  [/TR]
  [TR]
   [TD="class: xl65, align: right"]0.1[/TD]
   [TD="class: xl65, align: right"]6.463778[/TD]
   [TD="class: xl65, align: right"]-0.1[/TD]
   [TD="class: xl65, align: right"][COLOR=#ffffff]...............[/COLOR]8.768361
[/TD]
  [/TR]
  [TR]
   [TD="class: xl65, align: right"]0.01[/TD]
   [TD="class: xl65, align: right"]7.280365[/TD]
   [TD="class: xl65, align: right"]-0.01[/TD]
   [TD="class: xl65, align: right"]7.502121
[/TD]
  [/TR]
  [TR]
   [TD="class: xl65, align: right"]0.001[/TD]
   [TD="class: xl65, align: right"]7.377994[/TD]
   [TD="class: xl65, align: right"]-0.001[/TD]
   [TD="class: xl65, align: right"]7.400162[/TD]
  [/TR]
  [TR]
   [TD="class: xl65, align: right"]0.0001[/TD]
   [TD="class: xl65, align: right"]7.387948[/TD]
   [TD="class: xl65, align: right"]-0.0001[/TD]
   [TD="class: xl65, align: right"]7.390165[/TD]
  [/TR]
  [TR]
   [TD="class: xl65, align: right"]0.00001[/TD]
   [TD="class: xl65, align: right"]7.388945[/TD]
   [TD="class: xl65, align: right"]-0.00001[/TD]
   [TD="class: xl65, align: right"]7.389167[/TD]
  [/TR]
  [TR]
   [TD="class: xl65, align: right"]0.000001[/TD]
   [TD="class: xl65, align: right"]7.389045[/TD]
   [TD="class: xl65, align: right"]-1E-06[/TD]
   [TD="class: xl65, align: right"]7.389067[/TD]
  [/TR]
 [/TABLE]

e^2 = 7.389056