Limits to Infinity: Exponential Function

[BloodyFrozen]

Four is killing me...

Confunction identities? <- nvm don't think that's it

 

[micromass]

multiply numerator and denominator by cos(x)+1.

 

[BloodyFrozen]

Ohhhhhhh, I tried that, but never thought about plugging in 0...fml

sin(x)/(cos(x)+1) ->

0/2=0

....wow (again)

 

[NeroKid]

solution to 2 without using hospital or Taylor:
limx→inf (1+1/x)^x = e
<=> limx→inf xln(1+1/x) =1
<=> limx→0 ln(1+x)/x = 1
<=> limx→0 x - ln(1+x) = 0
<=> limx→0 (ex-1)/x = e^0 =1

 

[BloodyFrozen]

solution to 2 without using hospital or Taylor:
limx→inf (1+1/x)^x = e
<=> limx→inf xln(1+1/x) =1
<=> limx→0 ln(1+x)/x = 1
<=> limx→0 x - ln(1+x) = 0
<=> limx→0 (ex-1)/x = e^0 =1

I'm not quite following you. I know that (1+1/n)ⁿ = e as n->infinity

 

[NeroKid]

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

 

[BloodyFrozen]

How did you get from here:
<=> limx→inf xln(1+1/x) =1
to here?:
<=> limx→0 ln(1+x)/x = 1

 

[NeroKid]

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

 

[BloodyFrozen]

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

 

[BloodyFrozen]

Lastly,

How did you get these two steps?

<=> limx→0 ln(1+x')/x' = 1
<=> limx→0 x - ln(1+x) = 0

 

[NeroKid]

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

 

[BloodyFrozen]

Got it. Thanks