# Math Help - L'hospitals rule problems

1. ## L'hospitals rule problems

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. 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

3. Originally Posted by sbcd90
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
Thanks for the answer, but the answer to number 2 is 1. Idk how to solve it

4. I want to know if in this case $
\mathop {\lim }\limits_{x \to 0} \frac{{\sec ^2 x}}
{{\tan x}}
$
is correct to use lhopital rule, because evaluating, whe have: $\dfrac{1}{0}$

I mean, which are the indeterminate form, whereat we can use lhopital rule?

I only know, in fraction: $\dfrac{0}{0}$, $\dfrac{\infty}{\infty}$

Thanks

5. Originally Posted by Nacho
I want to know if in this case $
\mathop {\lim }\limits_{x \to 0} \frac{{\sec ^2 x}}
{{\tan x}}
$
is correct to use l`hopital rule, because evaluating, whe have: $\dfrac{1}{0}$
No. $\frac10$ 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

$\frac00,\,\frac\infty\infty,\,\frac{-\infty}\infty,\,\frac\infty{-\infty},\text{ and }\frac{-\infty}{-\infty}.$

6. Very good point and a common mistake when applying this rule.

7. Then:

$
\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)}}
$

$
\sin \left( {2x} \right) \leqslant 2x \Rightarrow \frac{1}
{{2x}} \leqslant \frac{1}
{{\sin \left( {2x} \right)}}
$

$
\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 $
\mathop {\lim }\limits_{x \to 0^ + } \left( {\tan x} \right)^{\frac{1}
{x}} = \infty
$

8. 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