I would appreciate any help on these problems. I'm studying for a test and I cant figure these out.
1. lim cos(r)/ln(r)
as r->infinity
2. lim (tan(x))^(1/X)
as x->0+
Thank You!
2>p=(tan x)^1/x
or, ln p=(1/x)* ln tan x
or,lt(x-->0+) ln p=lt(x-->0+)[ln tan x]/x
or,ln lt(x-->0+)p=lt(x-->0+)sec^2 x/tan x
=lt(x-->0+)[2*sec^2 x* tan x]/sec^2 x
=lt(x-->0+)[2*tan x]=0
lt(x-->0+)p=e^0=1.
1>lt(r--->infinity)cos r/ln r.=lt(r--->infinity)(a term that lies between -1 & 1)/ln r
approx=c/infinity=0
substitute y=(tan x)^(1/x)
lny=(1/x)ln(tanx)
Lets focus on the right side
lim ln(tanx)/x
as x-> 0+
The answer is 0/0, so you can use L'hospitals rule
lim sec(x))^2/(tan x)
as x-> 0+
lim (sin x)/(cos x)^3
as x-> 0+
equals 0/1=0
Now focus on the left side
lim lny = 0
as x-> 0+
lny=0
y=1. And that's your final answer
I'm still not sure how to solve the first one though. Please help!