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Math Help - L'hospitals rule problems

  1. #1
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    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!
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  2. #2
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    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
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  3. #3
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    Quote Originally Posted by sbcd90 View Post
    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
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  4. #4
    Member Nacho's Avatar
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    I want to know if in this case <br />
\mathop {\lim }\limits_{x \to 0} \frac{{\sec ^2 x}}<br />
{{\tan x}}<br />
is correct to use l`hopital rule, because evaluating, whe have: \dfrac{1}{0}

    I mean, which are the indeterminate form, whereat we can use l`hopital rule?

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

    Thanks
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  5. #5
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    Quote Originally Posted by Nacho View Post
    I want to know if in this case <br />
\mathop {\lim }\limits_{x \to 0} \frac{{\sec ^2 x}}<br />
{{\tan x}}<br />
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'H˘pital's rule are

    \frac00,\,\frac\infty\infty,\,\frac{-\infty}\infty,\,\frac\infty{-\infty},\text{ and }\frac{-\infty}{-\infty}.
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  6. #6
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    Very good point and a common mistake when applying this rule.
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  7. #7
    Member Nacho's Avatar
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    Then:

    <br />
\mathop {\lim }\limits_{x \to 0^ +  } \frac{{\sec ^2 x}}<br />
{{\tan x}} = \mathop {\lim }\limits_{x \to 0^ +  } \frac{1}<br />
{{\sin x\cos x}} = 2\mathop {\lim }\limits_{x \to 0^ +  } \frac{1}<br />
{{\sin \left( {2x} \right)}}<br />

    <br />
\sin \left( {2x} \right) \leqslant 2x \Rightarrow \frac{1}<br />
{{2x}} \leqslant \frac{1}<br />
{{\sin \left( {2x} \right)}}<br />

    <br />
\mathop {\lim }\limits_{x \to 0^ +  } \frac{1}<br />
{{2x}} = \infty  \Rightarrow \mathop {\lim }\limits_{x \to 0^ +  } \frac{1}<br />
{{\sin \left( {2x} \right)}} = \infty  = \mathop {2\lim }\limits_{x \to 0^ +  } \frac{1}<br />
{{\sin \left( {2x} \right)}}<br />

    Hence <br />
\mathop {\lim }\limits_{x \to 0^ +  } \left( {\tan x} \right)^{\frac{1}<br />
{x}}  = \infty <br />
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  8. #8
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    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!
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