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
I want to know if in this case $\displaystyle
\mathop {\lim }\limits_{x \to 0} \frac{{\sec ^2 x}}
{{\tan x}}
$ is correct to use l`hopital rule, because evaluating, whe have: $\displaystyle \dfrac{1}{0}$
I mean, which are the indeterminate form, whereat we can use l`hopital rule?
I only know, in fraction: $\displaystyle \dfrac{0}{0}$, $\displaystyle \dfrac{\infty}{\infty}$
Thanks
No. $\displaystyle \frac10$ is not an indeterminate form. Such a result means that the limit is infinity (and thus, does not exist). The indeterminate forms for which you may apply L'Hôpital's rule are
$\displaystyle \frac00,\,\frac\infty\infty,\,\frac{-\infty}\infty,\,\frac\infty{-\infty},\text{ and }\frac{-\infty}{-\infty}.$
Then:
$\displaystyle
\mathop {\lim }\limits_{x \to 0^ + } \frac{{\sec ^2 x}}
{{\tan x}} = \mathop {\lim }\limits_{x \to 0^ + } \frac{1}
{{\sin x\cos x}} = 2\mathop {\lim }\limits_{x \to 0^ + } \frac{1}
{{\sin \left( {2x} \right)}}
$
$\displaystyle
\sin \left( {2x} \right) \leqslant 2x \Rightarrow \frac{1}
{{2x}} \leqslant \frac{1}
{{\sin \left( {2x} \right)}}
$
$\displaystyle
\mathop {\lim }\limits_{x \to 0^ + } \frac{1}
{{2x}} = \infty \Rightarrow \mathop {\lim }\limits_{x \to 0^ + } \frac{1}
{{\sin \left( {2x} \right)}} = \infty = \mathop {2\lim }\limits_{x \to 0^ + } \frac{1}
{{\sin \left( {2x} \right)}}
$
Hence $\displaystyle
\mathop {\lim }\limits_{x \to 0^ + } \left( {\tan x} \right)^{\frac{1}
{x}} = \infty
$
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!