Limits to Infinity: Exponential Function

In summary: I don't think I'm supposed to use L'Hopital's Rule for these, but I can't think of any other ways to find the limits. Can anyone help?In summary, we can use L'Hopital's Rule to find the limits of e^(x-1)/x, cos(x)-1/sin(x), and other similar functions. However, for functions such as (x+sqrt(x^2+3))/(-2x-s
  • #36
u take the ln at the both side of limx --> inf (1+1/x)^x = e
and from there u switch to x' which equal to 1/x
 
Mathematics news on Phys.org
  • #37
How did you get from here:
<=> limx→inf xln(1+1/x) =1
to here?:
<=> limx→0 ln(1+x)/x = 1
 
  • #38
limx→inf xln(1+1/x) =1
<=> lim(1/x)→0 xln(1+1/x)=1
call x' = 1/x
limx'→0 ln(1+x')/x' =1
 
  • #39
NeroKid said:
limx→inf xln(1+1/x) =1
<=> lim(1/x)→0 xln(1+1/x)=1
call x' = 1/x
limx'→0 ln(1+x')/x' =1

Ok, I see.
Thanks
 
  • #40
Lastly,

How did you get these two steps?

<=> limx→0 ln(1+x')/x' = 1
<=> limx→0 x - ln(1+x) = 0
 
Last edited:
  • #41
because limx→0 ln(1+x)/x = 1 and limx→0 ln(1+x) = 0 and limx→0 x =0
so limx→0 ln(1+x) = limx→0 x
 
  • #42
Got it. Thanks
 

Similar threads

Replies
11
Views
2K
Replies
15
Views
3K
Replies
5
Views
2K
Replies
10
Views
1K
Replies
6
Views
3K
Replies
9
Views
569
Replies
7
Views
2K
Back
Top